HYDRAULICS 


BOOKS  BY 
R.  L.  DAUGHERTY 


HYDRAULIC  TURBINES 
Second  Edition 

190  pages,  6X9,  Illustrated $2.00 

CENTRIFUGAL  PUMPS 

192  pages,  6X9,  Illustrated $2.00 

HYDRAULICS 

267  pages,  6X9,  Illustrated $2.50 


HYDRAULICS 


BY 

R.  L.  DAUGHERTY,  A.  B.,  M.  E. 

ASSISTANT  PROFESSOR  OF  HYDRAULICS,  8IBLEY  COLLEGE, 

CORNELL  UNIVERSITY;    AUTHOR  OF  "HYDRAULIC 

TURBINES"    AND    "CENTRIFUGAL    PUMPS" 


FIRST  EDITION 


McGRAW-HILL  BOOK  COMPANY,  INC. 
239  WEST  39TH  STREET,  NEW  YORK 

6  AND  8  BOUVERIE  STREET,  LONDON,  E.  C. 

1916 


COPYRIGHT,  1916,  BY  THE 
McGRAW-HiLL  BOOK  COMPANY  INC. 


THE. MAPLE. PEESS.YORK.PA 


PREFACE 

This  book  has  been  prepared  as  a  text  for  students  who  are 
required  to  cover  a  wide  field  in  hydraulics  in  a  limited  amount  of 
time.  Therefore  the  treatment  has  been  made  as  brief  and  con- 
cise as  is  consistent  with  clearness.  Attention  has  been  given 
mostly  to  matters  which  are  of  fundamental  importance  and 
but  little  space  has  been  devoted  to  those  things  which  are  of 
small  practical  value,  except  where  necessary  to  illustrate  basic 
principles.  As  a  step  in  saving  the  student's  time  a  liberal  use 
has  been  made  of  diagrams,  curves,  and  half-tones.  These 
not  only  save  words  but  often  give  a  clearer  idea  at  a  glance  than 
can  be  obtained  in  any  other  way. . 

The  treatment  throughout  has  been  made  as  consistent  as  is 
possible.  The  solution  of  all  problems  involving  the  flow  of 
water  is  made  to  depend  upon  applications  of  Bernoulli's  theorem, 
which  is  the  key  to  a  rational  treatment  of  the  subject.  The 
student  is  not  told  in  the  very  beginning  that  V  =  V^gh  and 
then  compelled  to  unlearn  it  later.  Experience  in  the  class  room 
has  shown  that  many  students  will  persistently  apply  that 
formula  whether  it  fits  the  case  or  not.  By  deriving  it  at  a 
later  time  by  an  application  of  Bernoulli's  theorem,  they  will 
more  readily  see  that  it  is  a  very  special  case  and  thus  realize 
more  fully  its  limitations. 

An  effort  has  been  made  to  avoid  special  cases  so  far  as  is 
possible.  The  treatment  in  the  text  and  the  equations  are  for 
the  most  part  perfectly  general.  Special  cases  are  given  only 
when  necessary  to  illustrate  the  application  of  some  general 
principle,  or  where  a  special  case  makes  some  proposition  clearer, 
and  when  the  general  treatment  is  too  complex.  But  the  at- 
tention of  the  reader  is  called  to  the  fact  that  the  equations  there 
given  are  not  universally  applicable. 

Class-room  experience  has  shown  that  very  few  students  ob- 
tain a  true  physical  conception  of  the  subject  of  hydraulics.  To 
most  of  them,  even  some  of  the  best,  it  is  very  largely  an  abstract 
subject.  This  is  partly  due  to  the  fact  that,  with  their  limited 
experience  and  observation,  they  have  actually  seen  but  few  of 

v 

337860 


vi  PREFACE 

the  things  with  which  the  book  deals  and  hence  they  can  form 
no  adequate  mental  picture  of  the  physical  facts.  In  order  to 
overcome  this,  so  far  as  possible,  a  large  number  of  illustrations 
from  photographs  have  been  employed.  As  a  further  step  in 
implanting  a  true  physical  idea  in  the  mind  of  the  student,  a 
great  deal  of  care  has  been  exercised  in  the  arrangement  and  pre- 
sentation of  the  subject  and  a  constant  attempt  has  been  made 
to  connect  one  part  with  another.  In  many  cases  the  problems 
have  been  taken  from  actual  practice  and  have  also  been 
arranged  so  as  to  be  instructive  in  themselves. 

In  considering  turbines  and  centrifugal  pumps  the  first  es- 
sential is  to  convey  a  fair  idea  as  to  the  general  appearance, 
construction,  and  arrangement  of  such  machines  and  possibly 
some  simple  features  of  their  operation,  since  it  is  useless  to 
plunge  directly  into  a  mass  of  equations  which  are  no  more  than 
mathematical  gymnastics  to  most  students.  The  second  step 
should  be  the  presentation  of  the  principles  of  operation  together 
with  a  general  idea  as  to  actual  characteristics.  These  facts 
could  then  be  explained  by  as  much  theory  as  one  had  time  to 
go  into.  In  this  text  but  very  little  theory  has  been  given  and 
that  of  the  simplest  kind,  though  it  is  believed  that  what  is  given 
is  both  general  and  rational.  By  the  aid  of  this  theory  the  nature 
of  the  characteristics  of  these  machines  can  be  accounted  for. 
After  that  one  is  ready  to  take  up  certain  very  useful  and  prac- 
tical commercial  factors  by  the  aid  of  which  one  can  classify 
turbines  or  pumps,  can  compare  one  type  with  another,  and  can 
make  an  intelligent  selection  of  the  best  type  for  certain 
conditions. 

The  simple  theory  of  hydraulic  machinery  that  has  been  given 
here  covers  about  all  that  is  really  useful  in  a  text  of  this  scope. 
The  design  of  turbines  and  pumps  is  too  empirical,  and  requires 
too  much  judgment  and  experience  backed  up  by  a  good  supply 
of  test  data,  to  be  expressed  by  a  few  equations.  Any  brief 
treatment  of  this  phase  of  the  subject  would  be  false  and  mis- 
leading, hence  it  has  been  omitted.  For  any  more  extended 
treatment  of  these  subjects  the  reader  is  referred  to  other  publi- 
cations of  the  author. 

The  main  idea  underlying  the  entire  text  has  been  to  present 
fundamental  principles.  After  this  ground  has  once  been  cov- 
ered, those  who  desire  to  specialize  in  hydraulics  are  prepared  to 


PREFACE  vii 

study  certain  topics  more  intensively.  The  devotion  of  con- 
siderable space  to  an  account  of  experiments  and  test  data  is 
unwarranted  here,  though  the  student  should  not  lose  sight  of 
the  fact  that  the  study  of  such  is  desirable  when  important  work 
is  undertaken.  However,  a  sufficient  amount  of  information 
on  experimental  coefficients  and  empirical  factors  has  been  given 
so  that  a  correct  idea  may  be  formed  both  as  to  the  range  of 
values  and  the  considerations  that  enter  into  the  choice  of  a 
suitable  value  for  a  given  case. 

Very  naturally  some  very  important  topics  in  practical  hydrau- 
lics have  been  omitted  altogether  or  treated  very  briefly  and 
superficially  because  they  did  not  involve  fundamental  principles 
and  hence  were  not  within  the  scope  of  this  text,  or  else  were  of 
such  a  nature  as  to  belong  to  advanced  treatises.  The  final 
apology  which  the  author  makes  for  this  work  is  that  it  has  been 
prepared  primarily  to  meet  the  needs  of  his  own  classes. 

The  author  wishes  to  acknowledge  his  indebtedness  to  the 
various  parties  whose  names  are  attached  to  certain  of  the 
illustrations  for  their  kindness  in  furnishing  the  same.  He  is  also 
indebted  to  E.  H.  Wood,  Professor  of  Mechanics  of  Engineering 
in  Sibley  College,  and  to  D.  R.  Francis,  Instructor  in  Sibley 
College,  for  valuable  assistance  in  the  criticism  of  the  manu- 
script and  the  reading  of  the  proof.  R.  L.  D. 

ITHACA,  N.  Y. 
April,  1916. 


CONTENTS 

PAGE 

PREFACE v 

NOTATION xiii 

CHAPTER  I 

INTRODUCTION      1 

Definition  of  subject — Distinction  between  a  solid  and  a  fluid — 
Distinction  between  a  gas  and  a  liquid — Compressibility  of  water — 
Density  of  water — Accuracy  of  computations — Notation — Units — 
Problems. 

CHAPTER  II 

INTENSITY  OF  PRESSURE 7 

Definition  of  intensity  of  pressure — Variation  of  pressure  in  a 
liquid — Surface  of  equal  pressure — Pressure  the  same  in  all  direc- 
tions— Pressure  expressed  in  height  of  liquid — Barometer — 
Vacuum — Absolute  and  relative  or  gage  pressure — Instruments 
for  measuring  pressure — The  hydraulic  press — Problems. 

CHAPTER  III 

HYDROSTATIC  PRESSURE  ON  AREAS 19 

Total  pressure  on  plane  area — Depth  of  the  center  of  pressure — 
Lateral  location  of  center  of  pressure — Resultant  thrust  on  plane 
areas — Horizontal  pressure  on  curved  surface — Vertical  pressure 
on  curved  surface — Component  of  pressure  in  any  direction — 
Resultant  pressure  on  curved  surface — Pipes  under  pressure — 
Buoyant  force  of  the  water  and  flotation — Metacenter. 

CHAPTER  IV 

APPLICATIONS  OF  HYDROSTATICS      32 

The  gravity  dam — The  framed  dam — The  arch  dam — The  earth 
dam — Additional  notes  on  dams — Flashboards — Problems. 

CHAPTER  V 

HYDROKINETICS 44 

Actual  and  ideal  conditions — Critical  velocity — Steady  flow — 
Rate  of  discharge — Equation  of  continuity — General  equation  for 
steady  flow — Use  of  the  work  "head" — Energy  and  power  mean- 

ix 


x  CONTENTS 

ing  of  head — Correct  and  incorrect  applications  of  Bernoulli's 
theorem — Applications  of  general  equation — Problems. 

CHAPTER  VI 

APPLICATIONS  OF  HYDROKINETICS 62 

Definition  of  a  jet — Jet  coefficients — Flow  through  orifice — 
Orifice  in  case  of  unequal  pressures — Submerged  orifice — Values  of 
orifice  coefficients — Flow  through  short  tubes — Flow  through 
nozzles — Efficiency  of  nozzle — Venturi  meter — Large  vertical 
orifice — Weir — The  triangular  weir — The  rectangular  weir — The 
Francis  weir  formula — The  Bazin  weir  formula — Comments  on 
weirs — The  Cippoletti  weir — Special  weirs — The  Pitot  tube — The 
current  meter — Comments  on  Measurement  of  Water — Discharge 
under  varying  head — Problems. 

CHAPTER  VII 

FLOW   THROUGH   PlPES 96 

Loss  of  head  in  pipe  friction — Loss  of  head  at  entrance — Loss  of 
head  at  discharge — Loss  of  head  in  nozzle — Other  minor  losses  of 
head — Flow  through  long  pipe — Hydraulic  gradient — Effect  of  air 
at  summit — Hydraulic  slope — Chezy's  formula — Other  formulas 
for  pipe  friction — Values  of  friction  factors — Size  of  pipe  for  given 
discharge — Power  delivered  by  a  pipe — Pipe  line  with  pump — Pipe 
line  with  turbine — Construction  of  pipe  lines — Problems. 

CHAPTER  VIII 

UNIFORM  FLOW  IN  OPEN  CHANNELS 125 

Open  channels — Uniform  flow — Hydraulic  gradient — Equation 
for  uniform  flow — Kutter's  formula  for  c — Manning's  formula  for 
c — Bazin's  formula  for  c — Construction  of  open  channels — Non- 
uniform  flow  in  open  channels — Stream  gaging — Rating  curve — 
Problems. 

CHAPTER  IX 

HYDRODYNAMICS 141 

Dynamic  force  exerted  by  a  stream — Force  exerted  upon  pipe — 
Theory  6f  Pitot  tube — Water  hammer  and  surges  in  unsteady  flow 
— Relation  between  absolute  and  relative  velocities — Impulse  of  a 
jet — Reaction  of  a  jet — Distinction  between  an  impulse  and  a 
reaction  turbine — Theorem  of  angular  momentum — Torque  ex- 
erted upon  turbine  by  water — Torque  exerted  upon  water  by  cen- 
trifugal pump — Power — Definitions  of  heads — Definitions  of  tur- 
bine efficiencies — Definitions  of  pump  efficiencies — Centrifugal 
action  or  forced  vortex — Free  vortex — Problems. 


CONTENTS  xi 

CHAPTER  X 

DESCRIPTION  OF  THE  IMPULSE  WHEEL 165 

The  impulse  wheel — Buckets — Nozzles  and  governing — Condi- 
tions of  service. 

CHAPTER  XI 

DESCRIPTION  OF  THE  REACTION  TURBINE 177 

The  reaction  turbine — Runners — Gates  and  governing — The  draft 
tube — Cases  and  settings — Conditions  of  service. 

•CHAPTER  XII 

WATER  POWER  PLANTS 198 

Elements  of  a  water  power  plant — High  head  plants — Low  head 
plants. 

CHAPTER  XIII 

THEORY  OF  THE  IMPULSE  WHEEL 209 

Action  of  the  water — Force  exerted  by  jet — Power  of  wheel — 
Speed — Head  on  impulse  wheel — Problems. 

CHAPTER  XIV 

THEORY  OF  THE  REACTION  TURBINE 221 

Introductory  illustration — Torque  exerted — Power — Speed — 
Values  of  ce  and  <j>e  for  maximum  efficiency — Theory  of  the  draft 
tube — Head  on  reaction  turbine — Problems. 

CH  AFTER  ^XV 

TURBINE  LAWS  AND  FACTORS 231 

Operation  under  different  heads — Different  sizes  of  runner — 
Specific  speed — Uses  of  specific  speed — Factors  affecting  effi- 
ciency— Problems. 

CHAPTER  XVI 

THE  CENTRIFUGAL  PUMP      239 

Definition — Classification — Description  of  the  centrifugal  pump — 
Conditions  of  service — Head  developed — Measurement  of  head — 
Head  imparted  by  impeller— Centrifugal  pump  factors— Specific 
speed — Operation  at  different  speeds— Factors  affecting  efficiency 
— Problems. 

APPENDIX      262 

Table  of  areas  of  circles — Table  of  standard  pipe  sizes — Table  of 
numbers  to  the  %  power — Table  of  numbers  to  the  %  power — 
Fundamental  trigonometry. 

INDEX .  265 


NOTATION 

A    =  angle  between  V  and  u  (Fig.  142) 

a     =  angle  between  v  and  u  (Fig.  142) 

c      =  coefficient  of  discharge  or  coefficient  of  flow 

ce     =  coefficient  of  contraction 

cv    =  coefficient  of  velocity 

D    =  diameter  of  turbine  runner  or  pump  impeller  in  inches 

d     =  diameter  of  pipe  in  feet    . 

d"  =  diameter  of  pipe  in  inches 

e      =  efficiency 

€h    =  hydraulic  efficiency 

em    =  mechanical  efficiency 

ev    =  volumetric  efficiency 

F     =  area  in  square  feet;  in  turbines  and  pumps  it  is  the  total  area  of  the 

streams  measured  normal  to  the  absolute  velocity  of  the  water 
f      =  friction  factor  in  pipes 

=  area  in  square  feet  in  turbines  or  pumps  measured  normal  to  the 

relative  velocity  of  the  water 
G     =  any  weight  in  pounds 

g     =  acceleration  of  gravity  in  feet  per  second  per  second 
H    =  total  effective  head  in  feet,  =  p  +  z  +  V*/2g 
H'  =  any  loss  of  head  in  feet 
h     =  head  in  feet 
/     =  moment  of  inertia 
k     =  any  coefficient  of  loss 
I      =  any  length  in  feet 

m    =  hydraulic  mean  depth  (or  hydraulic  radius)  in  feet 
N    =  revolutions  per  minute 
Ns  =  specific  speed,  =  N  X  Vfi.hp./h*4 
n     =  factor  in  Kutter's  formula 

=  any  abstract  number 
P     —  total  pressure  or  force  in  pounds 
p     =  intensity  of  pressure  in  feet  of  water 
p'    =  intensity  of  pressure  in  pounds  per  square  foot 
p"  =  intensity  of  pressure  in  pounds  per  square  inch 
q     =  rate  of  discharge  in  cubic  feet  per  second 
r     =  radius  to  any  point  in  feet 
s      =  slope  of  hydraulic  gradient,  =  H'/l 

=  tangential  component  of  absolute  velocity,  =  V  cos  A 
T    =  torque  or  moment  of  a  force  in  foot  pounds 
u     =  linear  velocity  of  a  point  in  feet  per  second 

V    =  absolute  velocity  of  water  (or  relative  to  earth)  in  feet  per  second 

xiii 


xiv  ABBREVIATIONS 

v     —  velocity  of  water  relative  to  some  moving  point  in  feet  per  second 
W  =  pounds  of  water  per  second,  =  wq 
w     =  density  of  water  in  pounds  per  cubic  foot 

3     =  any  vertical  distance  in  feet;  in  measuring  "head"  it  is  a  vertical 
elevation  above  any  arbitrary  datum  plane 

<t>     =  ratio  of  peripheral  speed  of  turbine  runner  or  pump  impeller  to  \^2gh 
<f>e    =  value  of  0  for,  which  the  maximum  efficiency  is  obtained 
w     =  angular  velocity  in  radians  per  second,  =  2irN/QO  =  u/r 

Values  of  quantities  at  specific  points  will  be  indicated  by  subscripts. 
In  the  use  of  subscripts  (1)  and  (2)  the  water  is  always  assumed  to  flow 
from  (1)  to  (2). 

ABBREVIATIONS 

G.P.M.  =  gallons  per  minute 

Sec.  ft.    =  cubic  feet  per  second 

R.p.m.    =  revolutions  per  minute 

Hp.         =  horsepower 

B.Hp.     =  brake  horsepower  =  D.Hp. 

W.Hp.    =  water  horsepower 


HYDRAULICS 

CHAPTER  I 
INTRODUCTION 


1.  Definition  of  Subject. — Hydromechanics  is  the  science  of 
the  mechanics  of  fluids.  It  may  be  subdivided  into  three 
branches:  Hydrostatics  is  the  study  of  the  mechanics  of  fluids 
at  rest,  hydrokinetics  deals  with  the  flow  of  fluids,  while  hydro- 
dynamics is  concerned  with  the  forces  exerted  by  or  upon  fluids 
in  motion. 

Hydraulics  is  practical  hydromechanics,  that  is,  it  is  the  study 
of  the  applications  of  hydromechanics  to  engineering  problems.1 
While  it  might  deal  with  any  fluid  it  is  generally  restricted  to 
liquids  and  especially  to  water. 

By  idealizing  conditions  and  ignoring  phenomena  that  are 
known  to  exist,  it  is  possible  to  study  hydromechanics  as  a  subject 
in  pure  mathematics.  But  naturally  the  results  of  such  studies, 
though  interesting,  are  often  of  little  practical  value.  The 
determination  of  actual  results  by  rigorous  mathematics  is  often 
impossible  because  of  the  fact  that  the  exact  nature  of  certain 
hydraulic  phenomena  are  either  unknown  or  if  known  are 
so  complex  that  it  is  not  feasible  to  express  them  as  mathematical 
functions.  We  must,  therefore,  resort  to  a  combination  of  rigid 
mathematics,  empirical  expressions,  and  experimental  coeffi- 
cients. The  science  that  results,  based  partly  upon  pure  reason- 
ing and  partly  upon  experimental  evidence,  is  called  hydraulics. 

It  is  seen  that  hydraulics  is  not  an  exact  science.  In  its  actual 
applications  much  depends  upon  the  judgment  and  the  ex- 
perience of  the  engineer.  In  many  cases  it  is  necessary  to 
compute  or  estimate  results  for  which  satisfactory  experimental 
data  is  lacking.  And  in  applying  any  experimental  factors  or 
empirical  formulas  it  is  well  to  have  some  familiarity  with  the 

1  The  derivation  of  the  word  "hydraulics"  means  "flow  of  water  in  a 
pipe"  but  usage  has  given  the  word  a  much  broader  significance. 

1 


2  .  HYDRAULICS 

work  upon  which  they  were  based  in  order  to  judge  as  to  their 
application  to  the  case  in  hand. 

2.  Distinction  between  a  Solid  and  a  Fluid. — The  distinction 
between  a  solid  and  a  fluid  is  ordinarily  quite  clear  but  there  are 
plastic  solids  which  flow  under  the  proper  circumstances  and 
even  metals  may  flow  under  high  pressures.     On  the  other  hand, 
there  are  certain  very  viscous  liquids  which  do  not  flow  readily 
and  it  is  easy  to   confuse  them  with  the  plastic  solids.     The 
definition  of  a  fluid  as  a  substance  which  flows  must  be  extended 
therefore.     The  distinction  is  that  any  fluid,  no  matter  how  vis- 
cous, will  yield  in  time  to  the  slightest  stress.     But  a  solid,  no 
matter  how  plastic,  requires  a  certain  magnitude  of  stress  to  be 
exerted  before  it  will  flow. 

Also  when  the  shape  of  a  solid  is  altered  by  external  forces  the 
tangential  stresses  between  adjacent  particles  tend  to  restore  the 
body  to  its  original  figure.  With  a  fluid  these  tangential  stresses, 
which  are  proportional  to  the  viscosity,  can  act  only  while  the 
change  is  taking  place.  When  motion  ceases  the  tangential 
stresses  disappear  and  the  fluid  does  not  tend  to  regain  its 
original  shape. 

3.  Distinction  between  a  Gas  and  a  Liquid. — A  fluid  may  be 
either  a  gas  or  a  liquid.     A  gas  is  quite  compressible  and  when 
all  external  pressure  is  removed  it  tends  to  expand  indefinitely. 
A  gas  is,  therefore,  in  equilibrium  only  when  it  is  completely  en- 
closed.    A  liquid,  on  the  other  hand,  is  relatively  incompressible 
and  if  all  pressure,  except  that  of  its  own  vapor,  be  removed  the 
cohesion  between  adjacent  particles  holds  them  together  so  that 
the  liquid  does  not  expand  indefinitely.     Therefore,  a  liquid  may 
have  a  free  surface,  that  is,  a  surface  from  which  all  pressure  is 
removed,  except  that  of  its  own  vapor. 

The  volume  of  a  gas  is  greatly  affected  by  changes  in  either 
pressure  or  temperature  or  both.  It  is  usually  necessary,  there- 
fore, to  take  account  of  changes  in  volume  and  temperature  when 
dealing  with  gases.  Since  the  mechanics  of  gases  is  largely  one  of 
heat  phenomena  it  is  called  thermodynamics. 

The  volume  of  a  liquid  is  affected  to  a  very  small  extent  by 
changes  in  pressure  or  temperature  and  for  most  purposes  the 
changes  in  volume  or  temperature  may  be  ignored. 

4.  Compressibility  of  Water. — Water  is  usually  said  to  be 
incompressible  and  as  compared  with  gases  it  is  relatively  so. 
But  it  is  much  more  compressible  than  many  solids  such  as  steel 
or  even  wood  where  the  elastic  limit  is  not  passed.     Its  bulk  or 


INTRODUCTION 


volume  modulus  of  elasticity,  the  ratio  of  the  change  of  pressure 
per  unit  area  to  the  change  per  unit  of  volume,  is 

Ev  =  294,000  Ib.  per  sq.  in. 

This  value  holds  only  for  pressures  below  1,000  Ib.  per  sq.  in.  and 
for  temperatures  near  the  freezing  point.  For  higher  tempera- 
tures it  increases  slightly.  Thus  at  77°F.  it  is  about  327,000 
Ib.  per  sq.  in.  and  at  212°F.  it  is  360,000  Ib.  per  sq.  in.  Also  for 
higher  pressures  than  the  above  the  modulus  is  materially  larger. 
Thus  at  a  pressure  of  65,000  Ib.  per  sq.  in.  Hite  found  a  value 
of  Ev  =  650,000  Ib.  per  sq.  in. 


10 


20 


Temperature  _  Degrees  Centigrade 
30  40  50  60  70 


80 


100 


62.6 
62.4 
62.2 
.  62.0 

$«« 

1 

1 

l 

1 

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30    40     50      60      70     80     90     100    110    120    130   140    150    160    170    180    130    200    210 
Temperature -Degrees  Fahrenheit 

FIG.  1. — Density  of  pure  water. 

It  may  be  seen  that  increasing  the  pressure  from  atmospheric 
to  1,000  Ib.  per  sq.  in.  will  reduce  the  volume  of  a  body  of  water 
by  985  -r-  294,000  or  about  %  of  1.0  per  cent.  Therefore,  it  is 
seen  that  the  usual  assumption  regarding  water  as  incompressible 
is  justified. 

5.  Density  of  Water. — The  density  of  water  varies  somewhat 
with  the  temperature  as  well  as  with  the  pressure.  In  Fig.  1  can 
be  seen  the  values  of  the  density  of  pure  water  at  atmospheric 


4  HYDRAULICS 

pressure  for  the  range  of  temperature  from  freezing  to  boiling. 
The  presence  of  impurities  increases  these  values  somewhat. 
Thus  ocean  water  may  ordinarily  be  taken  as  weighing  64.0  Ib. 
per  cu.  ft.  In  the  computations  in  this  book  it  will  be  sufficient 
to  take 

w  =  62.4  Ib.  per  cu.  ft. 

6.  Accuracy  of  Computations. — No  computed  result  can  be 
more  accurate  than  the  data  upon  which  it  is  based  and  it  is 
therefore  not  only  useless  but  also,  misleading  to  carry  out  results 
to  more  significant  figures  than  the  data  warrant.     It  should 
be  noted  that  the  number  of  significant  figures  has  no  relation 
to   the   location    of   the   decimal   point.     Thus    347,000,    34.7, 
0.0000347  are  all  values  given  to  three  significant  figures  and 
are  of  the  same  degree  of  accuracy.     It  is  incorrect  in  such  a 
figure  as  the  first  to  preserve  any  more  figures  such  as  347,129 
if  three  figure  work  is  all  that  is  warranted.     A«nd  if  it  is  warranted 
it  is  likewise  incorrect  in  such  a  value  as  the  last  to  abbreviate  it 
to  0.00003  for  that  is  equivalent  to  saying  that  its  value  is 
0.0000300  to  three  significant  figures. 

There  are  some  quantities  that  may  be  known  with  a  high 
degree  of  accuracy  but  in  hydraulic  work  most  experimental 
factors  are  uncertain  in  the  third  significant  figure  and  there  are 
some  coefficients  or  values  which  are  uncertain  even  in  the  second 
significant  figure.  Thus  slide  rule  work  is  all  that  is  usually 
justified. 

Suppose  for  example  that  the  product  is  desired  of  two  quanti- 
ties whose  values  are  34.7  and  125.  Multiplying  these  two 
numbers  together  we  obtain  4,337.5  but  the  answer  that  should 
be  given  is  4,340.  If  our  values  were  known  to  be  34.700  and 
125.00  then  the  exact  product  would  be  permissible.  But  if  our 
values  are  experimental  they  may  range  for  example  from  34.6 
to  34.8  and  124  to  126  respectively.  The  products  of  the 
minimum  and  maximum  values  in  each  case  are  4,290  and  4,380, 
thus  showing  that  our  result  of  4,340  is  uncertain  in  the  third 
significant  figure  as  we  should  expect  when  the  given  data  are 
not  correct  in  the  third  figure. 

7.  Notation. — The  use  of  a  systematic  and  consistent  notation 
is  highly  desirable  and  familiarity  with  the  notation  will  save 
time  and  trouble.     A  table  of  the  notation  employed  in  this 
book  is  given  on  page  xiii. 


INTRODUCTION  5 

So  far  as  possible  an  attempt  has  been  made  to  employ  the 
same  notation  that  the  majority  of  other  writers  use  in  this  and 
in  related  subjects.  This  will  result  in  a  few  cases  of  the  same 
letter  being  used  for  different  quantities  but  in  such  instances  the 
quantities  are  so  unlike  that  it  is  believed  no  real  confusion  can 
result.  Unfortunately,  the  necessity  of  avoiding  real  conflicts 
in  notation  prevents  the  use  of  certain  letters  that  most 
naturally  suggest  themselves  for  several  different  quantities,  the 
quantities  not  being  sufficiently  removed  from  each  other  to 
permit  the  duplication. 

8.  Units. — The  standard  system  of  units  employed  in  this 
book  is  based  upon  the  foot,  pound,  and  second.  With  few 
exceptions  all  formulas  are  to  be  used  with  such  units.  There  are 
some  few  exceptions  that  commercial  practice  makes  necessary  or 
desirable.  For  instance  the  diameter  of  a  pipe  is  customarily 
given  in  inches  rather  than  in  feet.  Any  exceptions  to  the  general 
rule  will  be  clearly  indicated. 

It  should  be  noted  that  the  units  of  the  answer  in  any  computa- 
tion can  be  determined  from  the  units  involved  in  the  separate 
items.  Thus  the  product  of  velocity  and  area  is  the  product  of 
(ft. /sec.)  X  sq.  ft.  =  cu.  ft. /sec.  The  familiar  quantity  v2/2g  is 
(ft. /sec.)2 /(ft. /sec.2)  =  ft.  The  product  of  the  depth  of  water 
by  the  density  of  water  is  ft.  X  lb.  per  cu.  ft.  =  Ib.  per  sq.  ft.,  etc. 

It  will  frequently  be  necessary  to  use  the  value  of  g,  the  ac- 
celeration of  gravity.  Its  units  are  feet  per  second  per  second, 
often  written  ft. /sec.2  The  value  of  g  varies  with  latitude  and 
elevation.  Its  value  for  any  locality  may  be  computed  by  the 
following  formula  according  to  Pierce, 

g  =  32.0894  (1  +  0.0052375  sin2Z)(l  -  0.0000000957  e), 

where  I  is  the  latitude  in  degree  and  e  is  the  altitude  in  feet.     For 
ordinary  purposes  g  may  be  taken  as  32.2  ft.  per  sec.2 


9.  PROBLEMS 

1.  If  a  body  of  water  is  subjected  to  a  pressure  of  65,000  lb.  per  sq.  in. 
how  much  less  will  its  volume  be  than  in  a  perfect  vacuum? 

Ans.  10  per  cent. 

2.  What  pressure  will  be  required  to  reduce  the  volume  of  a  body  of 
water  by  0.1  of  1.0  per  cent,  if  the  temperature  is  32°F.  and  the  initial 
pressure  10  lb.  per  sq.  in. 

Ans.  304  lb.  per  sq.  in. 


6  HYDRAULICS 

3.  If  the  temperature  is  77°F.  what  would  be  the  result  in  problem  2? 
Ans.  337  Ib.  per  sq.  in. 

4.  A  cubic  foot  of  ocean  water  at  the  surface  and  at  ordinary  temperature 
weighs  64.0  Ib.     At  the  surface  it  is  under  a  pressure  of  14.7  Ib.  per  sq.  in. 
What  will  be  the  weight  of  a  cubic  foot  at  a  depth  such  that  the  pressure  is 
2,000  Ib.  per  sq.  in.?     Assume  Ev  =  310,000  Ib.  per  sq.  in.     (Density  is 
inversely  proportional  to  volume.) 

6.  The  radiator  of  an  automobile  holds  2.0  cu.  ft.  of  water.  It  is  filled 
with  water  at  50°F'.  After  the  engine  has  been  running  the  temperature  of 
the  water  is  180°F.  Assuming  no  loss  by  evaporation  or  otherwise  and 
neglecting  expansion  of  radiator,  how  much  water  will  have  run  out  the 
overflow? 

6.  If  we  multiply  cubic  feet  of  water  by  the  density  of  water  in  pounds 
per  cubic  foot  and  by  feet,  in  what  units  will  the  answer  be? 

7.  If  we  multiply  torque,  which  is  the  product  of  a  force  in  pounds  and  a 
distance  in  feet,  by  angular  velocity  in  radians  per  second,  what  units  will 
be  involved  in  the  answer? 

8.  If  we  multiply  pounds  per  second  by  feet  per  second  and  divide  by  g,  in 
what  units  is  the  answer? 

9.  If  we  multiply  a  force  in  pounds  by  velocity  in  feet  per  second  in  what 
units  is  the  answer? 


CHAPTER  II 


INTENSITY  OF  PRESSURE 

10.  Definition  of    Intensity  of    Pressure. — By    intensity   of 
pressure  is  meant  pressure  per  unit  area.     It  may  be  expressed 
in  various  units  such  as  pounds  per  square  inch,  pounds  per 
square  foot,  or,  as  will  be  seen  later,  in  feet  of  water,  inches  of 
mercury,  etc. 

If  P  represents  the  total  pressure  on  some  finite  area,  F,  while 
dP  represents  the  total  pressure  on  an  infinitesimal  portion  of 
area,  dF,  the  intensity  of  pressure  is 

"-£  (1) 

If  the  pressure  is  uniformly  distributed  over  the  area  in  question 
the  intensity  of  pressure  would  then  be  p'  =  P/F.  If  the  pres- 
sure is  not  uniformly  distrib- 
uted the  latter  expression  will 
give  the  average  value  only. 

The  word  "pressure"  is  usu- 
ally used  for  "intensity  of 
pressure"  though  the  latter 
term  should  be  employed 
where  there  is  any  possibility 
of  misunderstanding.  The 
word  " pressure"  is  also  used 
to  designate  the  resultant 

force  exerted  on  an  area.  In  order  to  clearly  distinguish  this 
usage  from  intensity  it  would  be  well  to  employ  the  term  "re- 
sultant pressure"  or  "total  pressure." 

11.  Variation  of  Pressure  in  a  Liquid. — Let  us  consider  a 
slender  prism  of  the  liquid  in  Fig.  2  as  a  free  body  in  equilibrium. 
The  forces  acting  upon  it  will  be  the  pressures  on  its  various  faces 
and  the  pull  of  gravity.     If  the  intensity  of  pressure  at  A  be 
denoted  by  p'i,  the  total  pressure  on  the  end  at  A  will  be  p\dF, 
where  dF  is  the  cross-section  area.     In  similar  manner  the  total 
pressure  on  the  end  at  B  will  be  p'zdF.     The  weight  of  the  volume 


8  HYDRAULICS 

of  liquid  is  evidently  wdFl.  Since  the  prism  of  water  is  in 
equilibrium  the  algebraic  sum  of  the  components  in  any  direction 
of  all  the  forces  acting  on  it  will  be  zero.  If  the  forces  be 
resolved  along  the  axis  AB  the  three  forces  mentioned  will  be  the 
only  ones  that  will  appear  since  the  forces  acting  on  the  sides 
of  the  prism  are  all  normal  to  the  axis.  Hence  we  may  write 

p'i  dF  -  p'z  dF  +  wdFl  cos  a  =  0. 
Since  I  cos  a.  =  zz  —  z\  it  follows  that 

p'z  —  p'i  =  w(zz  —  zi)  (2) 

This  equation  shows  that  the  difference  in  the  intensity  of 
pressure  at  two  different  points  varies  directly  as  the  difference 
in  the  depths  of  the  two  points.  Also  if  point  A  be  taken  at  the 
level  where  p'i  is  zero  and  if  z  be  the  elevation  of  such  level  above 
any  other  point,  then  in  general 

p'  =  wz  (3) 

From  this  equation  it  can  be  seen  that  the  intensity  of  pressure 
varies  directly  as  the  depth  of  the  point  in  question  below  the 
level  where  pr  is  zero. 

The  results  of  equations  (2)  and  (3)  are  strictly  true  only  for  an 
incompressible  fluid  in  which  the  density,  wt  is  constant  at  all 
depths.  For  practical  purposes  water  is  an  incompressible  fluid 
and  hence  these  equations  may  be  applied.  But,  owing  to  the 
high  degree  of  compressibility  of  gases,  they  should  not  be  used 
for  a  gas  except  where  there  are  relatively  small  differences  in 
pressure. 

12.  Surface  of  Equal  Pressure. — It  may  be  seen  from  equation 
(3)  that  all  points  in  a  connected  body  of  water  at  rest  are  under 
the  same  intensity  of  pressure  if  they  are  at  the  same  depth. 
This  indicates  that  a  surface  of  'equal  pressure  is  a  horizontal  plane. 
Strictly  speaking  it  is  a  surface  everywhere  normal  to  the  direc- 
tion of  gravity  and  it  is,  therefore,  approximately  a  spherical 
surface  concentric  with  the  earth.     But  a  limited  portion  of  such 
a  spherical  surface  is  practically  a  plane  area. 

A  free  surface  is  strictly  one  on  which  there  is  no  pressure. 
Usually  however  the  surface  of  a  liquid  exposed  only  to  the  pres- 
sure of  the  atmosphere  is  said  to  be  a  free  surface. 

13.  Pressure  the  Same  in  all  Directions. — In  a  solid,  owing  to 
the  existence  of  tangential  stresses  between  adjacent  particles, 


INTENSITY  OF  PRESSURE  9 

the  stresses  at  a  given  point  may  be  different  in  different  direc- 
tions. But  in  a  fluid  at  rest  no  tangential  stresses  can  exist  and 
the  only  forces  exerted  between  adjacent  surfaces  are  normal 
to  the  surfaces.  Therefore,  the  intensity  of  pressure  at  a  given 
point  is  the  same  in  every  direction. 

This  can  be  proven  by  reference  to  Fig.  3  where  we  have  a  small 
triangular  element  of  volume  whose  thickness  perpendicular  to  the 
plane  of  the  paper  is  constant  and  equal  to  dz.  Let  a  be  any 
angle,  p'  the  intensity  of  pressure  in  any  direction,  and  p'x  the 
intensity  of  pressure  on  a  vertical  plane.  The  following  forces 
act  upon  this  volume  of  fluid:  The 

, ,  ,  .       ,  j.  .         i   j    j  \  P.  dl  •  dz 

pressure  on  the  vertical  tace  is  px  ayaz, 

the   pressure   on  the  slanting  face  is 

p'dldz,  then  there  are  the  pressures  on 

the  horizontal  face  and  on  the  two 

faces    parallel    to    the    plane    of  the 

paper,  and  the  weight  of  the  volume. 

Their  values  are.  not  required.    Since  „ 

this  volume  is   a  fluid  body  at  rest 

there  are  no  other  forces  besides  these  normal  to  the  surfaces,  and, 

since  it  is  a  condition  of  equilibrium,  the  sum  of  the  components 

in  any  direction  is  equal  to  zero.     Writing  such  an  equation  for 

components  in  a  horizontal  direction  we  have  only 

p'dldz  cos  a  —  p'x  dydz  =  0. 
Since  dy  =  dl  cos  a,  it  is  seen  that 

P'  =  P'* 

This  result  is  independent  of  the  angle  a,  and,  therefore,  it  follows 
that  the  intensity  of  pressure  is  the  same  upon  any  plane  passing 
through  0. 

14.  Pressure  Expressed  in  Height  of  Liquid. — In  Fig.  4 
imagine  a  body  of  liquid  upon  whose  surface  there  is  no  pressure. 
Then  by  equation  (3)  the  intensity  of  pressure  at  any  depth  z 
is  p'  =  wz.  For  a  given  liquid  w  is  constant  and  thus  there  is 
a  definite  relation  between  p'  and  z.  That  is  any  pressure  per 
unit  area  is  equivalent  to  a  corresponding  height  of  liquid.  In 
hydraulic  work  it  is  often  more  convenient  to  express  intensity 
of  pressure  in  terms  of  height  of  a  column  of  water  rather  than  in 
pressure  per  unit  area. 

Even  if  the  surface  of  the  liquid  in  Fig.  4  is  under  some  pressure 


10  HYDRAULICS 

the  relation  stated  is  still  true.  For  this  pressure  on  the  surface 
could  be  expressed  in  terms  of  height  of  the  liquid  and  such  value 
added  to  z.  The  resulting  value  of  p'  would  thus  be  increased  by 
the  amount  of  this  surface  pressure. 

The  intensity  of  pressure  expressed  in  terms  of  the  height  of  a 
column  of  liquid  will  be  denoted  by  the  letter  p.     It  will  thus  be 

seen  that  we  have  the  relation 

p'  =  wp  (4) 

This  equation  is  true  for  any  con- 
sistent system  of  units.     Thus  if  w 
is  density  in  pounds  per  cubic  foot, 
p  must  be  expressed  in  feet,  and 
•pIG   4  p'  will  then  be  in  pounds  per  square 

foot.     For  pure  water  at  ordinary 

temperatures  we  have  the  relation  pf  =  62.4p.  It  is  quite  com- 
mon to  express  intensity  of  pressures  in  pounds  per  square  inch, 
but  p  is  rarely  found  expressed  in  other  units  besides  feet  of 
water.  Since  p'  =  144p"  =  62.4p  we  have 

p"  =  0.4333p  and  p  =  2.308p" 


EXAMPLES 

1.  Neglecting  the  pressure  of  the  atmosphere  upon  the  surface,  what  is 
the  pressure  in  pounds  per  square  inch  at  a  depth  of  3,000  ft.  in  fresh  water? 
At  a  depth  of  3,000  ft.  in  the  ocean? 

2.  A  certain  pump  for  a  hydraulic  press  delivers  water  at  a  pressure  of 
5,000  Ib.  per  sq.  in.     To  what  height  of  pure  water  would  that  be  equivalent? 
To  what  height  of  liquid  having  a  density  of  100  Ib.  per  cu.  ft.? 

3.  The  specific  gravity  of  mercury  is  13.57,  that  is,  its  density  is  13.57 
times  that  of  pure  water.     How  many  feet  of  mercury  is  equivalent  to  a 
pressure  of  100  Ib.  per  sq.  in.? 

4.  If  the  specific  gravity  of  mercury  is  13.57  how  many  feet  of  water  is 
equivalent  to  a  pressure  of  10  ft.  of  mercury? 

6.  The  pressure  of  the  atmosphere  is  about  14.7  Ib.  per  sq.  in.  To  what 
height  in  feet  of  water  is  this  equivalent?  What  is  the  equivalent  height 
in  inches  of  mercury? 

15.  Barometer. — If  a  tube  such  as  that  in  Fig.  5  has  its  lower 
end  immersed  in  a  liquid  and  the  air  is  exhausted  from  the  tube 
the  liquid  will  rise  in  the  latter.  If  the  air  is  completely  ex- 
hausted we  shall  have  zero  pressure  on  the  surface  of  the  liquid 
in  the  tube,  and  the  liquid  will  have  reached  its  maximum  height. 


INTENSITY  OF  PRESSURE 


11 


1 


This  device  is  called  the  barometer  and  is  used  for  measuring  the 
pressure  of  the  atmosphere. 

By  Art.  12  it  may  be  seen  that  the  intensities  of  pressure  at  o 
(within  the  tube)  and  at  a  (at  the  surface  of  the  liquid  outside) 
are  the  same.  That  is  p0  =  pa>  And,  since  the  pressure  on  the 
surface  of  the  liquid  in  the  tube  is  zero,  the  intensity  of  pressure 
at  o  is  by  equation  (3) 

p'o  =  wy. 

And  by  equation  (4)  p'0  =  wp0.     Thus  the  pressure  of  the  air  in 
terms  of  height  of  the  liquid  column  is 

Pa  =  y  (5) 

The  liquid  employed  is  usually  mercury  because  of  the  fact 
that  its  density  is  sufficiently  great  to  enable  a  reasonably 
short  tube  to  be  used,  and  because  its  vapor 
pressure  is  negligible  at  ordinary  tempera- 
tures. If  water  were  employed  the  height  of 
tube  would  be  inconvenient  and  also  its  vapor 
pressure  at  ordinary  temperatures  is  appreci- 
able so  that  instead  of  having  a  perfect  vac- 
uum at  the  top  of  the  tube  we  should  have  a 
space  filled  with  water  vapor.  The  height 
attained  by  the  liquid  would  consequently  be 
less  than  what  would  otherwise  be  the  case.  The  diameter  of 
the  tube  should  be  at  least  0.5  in.  in  order  to  eliminate  errors  due 
to  capillarity. 

The  pressure  of  the  atmosphere  is  different  in  different  locali- 
ties, depending  upon  elevation,  and  at  a  given  point  it  varies  from 
time  to  time  according  to  the  temperature  and  other  factors. 

In  round  numbers  the  pressure  of  the  atmosphere  may  be 
taken  as  14.7  Ib.  per  sq.  in.,  30  in.  of  mercury,  and  34  ft.  of  water. 
(These  values  are  not  exact  equivalents.) 

EXAMPLES 

1.  If  the  barometer  reads  29.92  in.  of  mercury,  what  is  the  pressure  in 
pounds  per  square  inch? 

2.  If  the  pressure  of  water  vapor  at  80°F.  is  0.505  Ib.  per  sq.  in.  what 
would  be  the  height  of  the  water  barometer  if  the  atmospheric  pressure  were 
14.7  Ib.  per  sq.  in.?     (Use  correct  density  of  water  for  this  temperature.) 

3.  Assuming  the  density  of  air  to  be  0.0807  Ib.  per  cu.  ft.,  what  would  be 
the  height  of  the  air  surrounding  the  earth  and  producing  a  pressure  of  14.7 
Ib.  per  sq.  in.,  if  air  were  incompressible? 


FIG.  5. — Barometer. 


12  HYDRAULICS 

16.  Vacuum. — Pressures  less  than  that  of  the  atmosphere  are 
usually  called  vacuums,  a  perfect  vacuum  meaning  an  entire 
absence  of  all  pressure.     Vacuum  is  usually  measured  from  the 
pressure  of  the  atmosphere  as  a  base  and  is  commonly,  though  not 
necessarily,  measured  in  inches  of  mercury.     If  the  atmospheric 
pressure  is  30  in.  of  mercury,  a  perfect  vacuum  would  then  be  a 
vacuum  of  30  in.     And  a  vacuum  of  10  in.  of  mercury  would 
mean  that  there  was  a  real  pressure  of  20  in.  of  mercury. 

EXAMPLES 

1.  If  the  barometer  reads  28.5  in.  of  mercury  and  the  absolute  pressure  in 
the  condenser  for  a  steam  turbine  is  1.5  in.  of  mercury,  what  is  the  value  of 
the  vacuum? 

2.  The  barometer  reads  30  in.  of  mercury  and  within  a  certain  vessel  there 
is  a  vacuum  of  22  in.  of  mercury.     What  fs  the  real  pressure  within  that 
vessel  in  pounds  per  square  inch?     What  is  the  excess  external  pressure  on 
the  walls  of  the  vessel  in  pounds  per  square  inch? 

17.  Absolute  and  Relative  or  Gage  Pressures. — If  the  pres- 
sure is  measured  above  absolute  zero  pressure  it  is  called  absolute 
pressure.     If  it  is  measured  from  the  atmospheric  pressure  as  a 

base  it  is  called  relative  or  gage  pressure,  since 
it  is  only  relative  pressure  that  a  gage  meas- 
ures. Thus  Fig.  6  shows  a  compound  gage 
which  will  measure  pressures  either  above  or 
below  that  of  the  atmosphere.  When  the 
gage  is  open  to  the  atmosphere  the  hand 
points  to  zero.  If  the  gage  is  connected  to 

,    any  vessel  in  which  there  is  a  pressure  above 
FIG.  6.— Compound      ,  J 

gage.  that  of  the  surrounding  air  the  hand  will  turn 

in  a  clockwise  direction  from  zero.  If  the  pres- 
sure is  a  vacuum  the  hand  will  move  in  the  opposite  direction. 
Thus  the  gage  measures  only  the  difference  between  the  pressure 
on  the  inside  of  the  gage  tube  and  that  of  the  air  surrounding  the 
gage. 

In  Fig.  7  let  0  indicate  entire  absence  of  all  pressure  or  absolute 
zero  and  the  ordinate  OA  represent  the  pressure  of  the  atmos- 
phere. Then  suppose  we  have  any  pressure  such  as  at  B.  The 
gage  will  read  the  value  AB  and  this  is  the  gage  pressure.  The 
absolute  pressure  is  OB.  Also  if  we  have  a  vacuum  of  AC,  the 
gage  pressure  is  —AC,  the  minus  sign  merely  indicating  a  value 
below  atmospheric  just  as  a  plus  sign  indicates  a  pressure  above 
that  of  the  atmosphere.  But  the  absolute  pressure  is  OC. 


.Absolute  Zero 


INTENSITY  OF  PRESSURE  13 

When  dealing  with  absolute  pressures  all  values  are  positive. 
In  the  case  of  gage  pressures  only  values  above  that  of  the  atmos- 
pheric pressure  are  positive,  but  the  minus  sign  for  pressures 
below  that  of  the  atmosphere  serves  only  to  indicate  a  vacuum. 
There  may  still  be  a  real  pressure  between 
adjacent  particles  of  water.  A  true  negative 
pressure  would  mean  that  the  water  was  in  a 
state  of  tension  and  as  water  can  sustain  only 
a  very  slight  tensile  stress  it  is  impossible  to 
have  a  pressure  below  absolute  zero.  Abso- 
lute zero  is  the  point  where  the  stress  in  the 
liquid  would  change  from  compression  to 
tension. 

In    most    problems   in   hydraulics   we   are     A  Atmospheric  Erc88UTe 
not  interested  in  absolute  pressures.     We  are     c 
concerned  with  the  differences  in  pressure  in- 
side   a  vessel  and  that  outside  for  example     o 
and  in  general  that  would  be  the  gage  pres-  FlG  7 

sure.     And  in  many  other  cases  the  atmos- 
pheric pressure  acts  alike  at  all  points  and  balances  out. 


EXAMPLES 

1.  A  gage  reads  20  Ib.  per  sq.  in.  when  the  gage  itself  is  surrounded  by  the 
atmosphere.     If  the  air  surrounding  the  gage  be  exhausted  to  a  vacuum  of 
20  in.  of  mercury  while  the  real  pressure  of  the  fluid  on  the  inside  of  the  gage 
tube  remains  the  same,  what  will  be  the  reading  of  the  gage? 

2.  If  the  barometer  reads  30  in.  of  mercury  and  a  vacuum  gage  reads  5  in. 
of  mercury,  what  is  the  absolute  pressure? 

3.  A  gage  pressure  of  —  25  ft.  of  water  is  how  much  less  than  a  gage  pres- 
sure of  10  ft.  of  water? 

18.  Instruments  for  Measuring  Pressure. — Gage. — The  famil- 
iar pressure  or  vacuum  gages  have  already  been  referred  to  and 
the  combination  of  the  two  is  shown  in  Fig.  6.  In  this  type 
of  instrument  a  curved  tube  is  caused  to  change  its  curvature 
by  changes  of  pressure  within  the  interior  of  the  tube.  The 
moving  end  of  the  tube  then  rotates  a  hand  by  means  of  some 
intermediate  links.  It  is  usually  assumed  that  the  pressure 
indicated  by  the  gage  is  that  existing  at  the  center  of  the  gage. 
Thus  the  location  of  the  center  of  the  gage  should  always  be 
taken  into  consideration.  For  instance,  referring  to  Fig.  8,  the 


14 


HYDRAULICS 


f 


pressure  at  A  is  the  gage  reading  plus  the  distance  z.     If  the  gage 
reads  pounds  per  square  inch,  as  is  customary, 

pA  =  2.308  p"  +  z. 

Piezometer  Tube. — A  piezometer  tube  is  a  simple  device  for 
measuring  moderate  pressures.  It  consists  of  a  tube  in  which  the 
liquid  can  freely  rise,  without  overflowing,  un- 
til equilibrium  is  established.  To  prevent  error 
due  to  capillarity  the  diameter  of  the  tube  should 
be  at  least  0.5  in.  The  height  of  the  surface  of 
liquid  in  the  tube  will  give  the  pressure  de- 
sired directly.  It  should  be  noted  that  if  the 
water,  whose  pressure  is  desired,  is  flowing  the 
true  pressure  can  be  obtained  only  by  having 
the  axis  of  the  tube  at  the  point  of  connection 
perpendicular  to  the  stream  flow  and  further- 
more the  interior  opening  should  be  smooth  and 
free  from  all  projections.  If  the  end  of  the  pipe 
projects  into  the  stream,  as  in  the  case  of  the  fourth  tube  in 
Fig.  94,  the  pressure  read  will  be  too  low. 

Mercury  U  Tube. — For  high  pressure  the  water  piezometer  is 
not  suited  and  some  modification  must  be  adopted.     The  mercury 


FIG.  8. 


FIG.  9. — Piezometer. 


FIG.  10. 


U  tube  shown  in  Fig.  10  may  then  be  used.  If  s  is  the  specific 
gravity  of  the  mercury  (or  whatever  liquid  may  be  employed)  the 
pressure  at  the  point  C  is  sy.  This  is  also  the  pressure  at  B 
but  the  pressure  at  A  is  greater  than  this  by  the  amount  z,  if  the 


INTENSITY  OF  PRESSURE 


15 


tube  from  A'  to  B  is  filled  with  water.  If  it  were  filled  with  air, 
then,  neglecting  the  slight  weight  of  the  air  within  the  tube,  the 
pressures  at  B  and  at  the  surface  of  the  water  at  A'  would  be 
equal.  In  practice  it  would  be  difficult  to  insure  the  absence  of 
air  and  if  the  tube  were  partially  filled  with  air  and  partially  with 
water  it  would  be  troublesome  to  make  correction  and  accuracy 
would  be  impossible  unless  it  were  known  just  what  proportion 
of  the  tube  was  filled  with  water  and  what  with  air.  It  is 
therefore  desirable  to  provide  some  means  of  permitting  all  the 
air  to  escape  and  its  place  to  be  taken  by  water.  If  the  con- 
necting tube  in  Fig.  10  is  filled  with  water  the  pressure  at  A  is 


PA 


+  sy- 


FIG.  11. 


FIG.  12. 


In  measuring  a  vacuum  we  must  interpret  y  as  a  negative 
quantity  in  Fig.  11  so  that  we  have,  if  the  tube  is  filled  with  water, 

pA  =  z  -  sij. 

If  this  connecting  tube  from  A'  to  B  were  filled  with  air  then 
the  correction  for  the  height  above  A'  would  be  negligible  but  it  is 
difficult  to  insure  this  being  filled  with  air  and  error  will  be  intro- 
duced if  it  is  not.  Thus  the  arrangement  in  Fig.  12  is  much  better 
as  that  permits  no  air  to  collect  in  the  tube  and  introduce  errors 
in  the  readings.  In  this  case  z  is  negative  so  that 


PA  =  -  z  - 


16 


HYDRAULICS 


Differential  Gage. — The  differential  gage  is  used  for  measuring 
differences  of  pressure  only.  One  form  of  this  is  shown  in  Fig.  13. 
Assuming  the  entire  connecting  tubing  to  be  filled  with  water 
except  that  portion  of  the  U  that  is  filled  with  the  denser  liquid, 
such  as  mercury  for  instance,  the  pressure  at  A'  will  exceed  that 
at  E'  by  the  amount  sy.  That  is 


But 
and 


FIG.  13. 

Substituting  these  values  we  have 

PA  ~  PB  =  W  +  ZA  -  ZB 

=  sy  -  y  =  (s  -  l)y. 

In  the  differential  manometer  the  left-hand  column  of  mercury, 
or  whatever  it  is,  has  a  column  of  water  of  height  y  resting  upon 
it  that  is  not  balanced  by  a  corresponding  amount  on  the  right- 
hand  column,  hence  the  pressure  difference  is  not  sy  alone. 

EXAMPLES 

1.  Two  pressure  gages  are  connected  to  the  same  vessel  containing  water 
under  pressure.     One  of  these  gages  is  10  ft.  below  the  point  where  the  pres- 


INTENSITY  OF  PRESSURE 


17 


sure  is  measured  and  it  reads  40  Ib.  per  sq.  in.  The  other  gage  is  located  15 
ft.  above  the  point  in  question.  What  will  it  read?  What  is  the  pressure  in 
the  vessel?  (It  is  assumed  that  the  connecting  pipes  are  filled  with  water.) 
2.  Two  vessels  are  connected  to  a  differential  manometer  using  mercury 
(specific  gravity  =  13.57).  When  the  mercury  reading  is  36  in.  what  is  the 
difference  in  pressure  in  feet  of  water  between  the  two  vessels? 

19.  The  Hydraulic  Press. — The  most  important  device  operat- 
ing upon  the  principle  of  equal  transmission  of  intensity  of  pres- 
sure in  all  directions  is  the  hydraulic  press.  If  in  Fig.  14  a 
force  PI  be  applied  to  the  small  piston  whose  area  is  FI  the 
intensity  of  pressure  throughout  the  whole  volume  of  liquid 
will  be  increased  by  the  amount  p'  =  Pi/Fi.  Then  the  total 
additional  force  exerted  upon  the  face  of  the  large  piston  will  be 
p'F2.  =  (Pi/Fi)Fz  =  Pi(F2/Fi).  It  is  thus  seen  that  a  small  force 


r 


r 


FIG.  14.  —  Hydraulic  press. 

exerted  on  the  smaller  piston  is  enabled  to  oppose  a  much 
greater  load  on  the  large  piston.  If  G\  and  Gz  denote  the 
weights  of  the  pistons  while  z  is  the  difference  in  elevation  of 
their  faces,  we  have  for  equilibrium 


»  . 

r  i  /<2 

Since  the  volume  of  liquid  in  the  vessel  must  remain  constant, 
it  follows  that  the  distance  moved  by  the  larger  piston  must  be 
much  less  than  that  moved  through  by  the  smaller  piston. 

EXAMPLES 

1.  In  Fig.  15  the  diameter  of  the  small  piston  is  %  in.  and  that  of  the  large 
one  is  20  in.     The  big  plunger  weighs  1,000  Ib.  and  sustains  an  external  load 
2 


18 


HYDRAULICS 


of  6,000  Ib.  The  liquid  used  is  water.  What  total  force  P  applied  to  the 
small  piston  will  secure  equilibrium? 

Ans.  4.12  Ib. 

2.  When  the  small  piston  has  descended  20  ft.  how  far  will  the  plunger 
have  been  raised  ? 

20.  PROBLEMS 

1.  In  Fig.  16  what  are  the  values  of  absolute  pressure  at  A,  B,  C,  and  D, 
assuming  the  liquid  to  be  water?  What  are  the  values  of  gage  pressure? 


T" 

•* 

D 

1                                                34/ 

'-.  ~- 

"T     •     I 

V 

•^ 

:X- 

--C 
A 

so'         floor         m 

-~- 

jg^^J=g| 

\           l            — 

§EE= 

^ 

teaa 

^-^  :  ' 

I     i^t 

P       J3                        v 

P^ 

FIG.  15. 


FIG.  16. 


FIG.  17. 


What  is  the  value  of  the  vacuum  at  C?     (Give  answers  in  feet  of  water, 
pounds  per  square  inch,  and  inches  of  mercury  in  every  case.) 

2.  In  Fig.  17  the  cylinder  is  2  ft.  in  diameter  and  the  weight  of  the  piston 
and  load  is  4,000  Ib.     What  will  be  the  gage  reading  in  pounds  per  square 
inch? 

3.  If  the  mercury  manometer  in  Fig.  17  reads  35  in.,  how  far  is  the  top  of 
the  lower  mercury  column  below  the  piston?     If  the  manometer  remained  at 


Open  Tube 


20,00<? 


2000* 

t  , 
10 
X 

Hr- 

FIG.  18. 


this  same  place  but  the  connection  were  made  to  the  tank  at  a  different 
level,  would  the  mercury  reading  change? 

4.  The  small  piston  in  Fig.  18  has  a  diameter  of  3  in.  Neglecting  friction, 
when  a  force  PI  of  20,000  Ib.  is  applied  to  it,  what  will  be  the  force  P2  that 
can  be  exerted  by  the  plunger  with  a  diameter  of  24  in.  ?  To  what  height 
would  water  rise  in  the  piezometer  tube  shown? 


CHAPTER  III 


Free  Surface 


HYDROSTATIC  PRESSURE  ON  AREAS 

21.  Total  Pressure  on  Plane  Area. — Since  we  are  dealing  with 
fluids  at  rest,  no  tangential  forces  can  be  exerted  and  hence  all 
pressures  are  normal  to  the  surfaces  in  question.  If  the  pressure 
were  uniformly  distributed  over  an  area,  the  total  or  resultant 
pressure  would  be  the  product  of  the  area  and  the  intensity  of 
pressure  and  the  point  of  application 
of  the  force  would  be  the  center  of 
gravity  of  the  area.  In  general  the  in- 
tensity of  pressure  is  not  uniform, 
hence  further  analysis  is  necessary. 

In  Fig.  19  consider  a  vertical  plane 
whose  upper  edge  lies  in  the  free  sur- 
face. Let  this  plane  be  perpendicular 
to  the  plane  of  the  paper  so  that  AB 
is  merely  its  trace.  The  intensity  of 
pressure  will  vary  from  zero  at  A  to 
BC  at  B.  It  will  thus  be  seen  that 
the  total  pressure  P  will  be  the  sum- 
mation of  the  products  of  the  ele- 
mentary areas  and  the  intensities  of 
pressure  upon  them.  It  is  also  ap- 
parent that  the  resultant  of  this  sys- 
tem of  parallel  forces  must  be  applied 

at  a  point  below  the  center  of  gravity  of  the  area,  since  the 
center  of  gravity  of  an  area  is  the  point  of  application  of  the  re- 
sultant of  a  system  of  uniform  parallel  forces.  If  the  plane  be 
immersed  to  A  'B'  the  intensity  of  pressure  varies  from  A'D  at  A' 
to  B'E  at  B'.  Since  the  proportionate  change  of  intensity  of 
pressure  from  A'  to  B'  is  less  than  before,  it  is  clear  that  the 
center  of  pressure  will  approach  nearer  the  center  of  gravity. 

In  Fig.  20  let  A  B  be  the  trace  of  a  plane  making  any  angle  B 
with  the  horizontal.  The  view  to  the  right  is  the  projection  of 
this  area  upon  a  vertical  plane  which  is  also  normal  to  the  plane 
containing  AB.  Let  z  be  the  depth  of  any  point  and  y  be  the 

19 


FIG.  19. 


20 


HYDRAULICS 


distance  of  any  point  from  OX,  the  axis  of  intersection  of  the 
plane,  produced  if  necessary,  and  the  free  surface.  The  coordin- 
ates of  the  center  of  gravity  of  the  area  may  be  denoted  by  z  and  y 
respectively. 

Take  an  element  of  area  dF  such  that  all  portions  of  said  ele- 
ment are  at  the  same  depth  z.     Then  the  total  pressure  on  dFis 

dP  =  p'dF  =  wzdF] 
Hence  P  =  wfzdF. 

But  fzdF  =  zFy 

hence  P  =  wzF  (6) 


Free  Surface 


22.  Depth  of  the  Center  of  Pressure.  —  The  point  of  applica- 
tion of  the  resultant  force  on  an  area  is  called  the  center  of  pres- 
sure. We  usually  locate  the  line  of  action  of  a  force  by  taking 
moments.  In  this  case  it  is  convenient  to  take  OX  in  Fig.  20 
as  the  axis  of  moments.  On  any  element  of  area  dF  the  total 
pressure  is 

dP  =  wzdF  =  wy  sin  ddF 
and  its  moment  is 

ydP  =  wy2  sin  BdF. 

If  the  distance  of  the  center  of  pressure  from  OX  be  denoted  by  y' 
fydP 


y' 


w  sn 


(7) 


w  sin  BfydF 

But  fy2dF  is  the  moment  of  inertia  of  the  area  F  about  the  axis 
OX,  and  fydF  is  the  statical  moment  of  the  area  with  respect  to 
the  same  axis,  hence 


This  may  be  put  in  a  more  convenient  form  by  noting  that,  if 
Ig  is  the  moment  of  inertia  of  the  plane  area  about  its  gravity 


HYDROSTATIC  PRESSURE  ON  AREAS  21 

axis  and  kg  is  the  radius  of  gyration  about  the  gravity  axis,  we 
have 


-  5  +  f  do) 

From  these  equations  it  may  be  seen  that  the  location  of  the 
center  of  pressure  is  independent  of  the  angle  0,  that  is,  the  plane 
area  may  be  rotated  about  the  axis  OX  without  affecting  the  loca- 
tion of  the  center  of  pressure.  However,  this  will  not  hold  for 
6  =  zero  since  the  value  of  P  would  also  be  zero. 

From  equation  (10)  it  may  also  be  seen  that  the  center  of 
pressure  is  always  below  the  center  of  gravity.  Also  as  the  depth 
of  immersion  is  increased  for  a  given  value  of  6,  the  distance 
y  increases.  But  as  kg  remains  constant  in  value  it  may  be 
seen  that  the  last  term  in  equation  (10)  becomes  relatively 
small,  hence  y'  approaches  y  in  value.  The  same  thing  would  be 
true  if  the  depth  of  the  center  of  gravity  z  remained  constant 
while  the  plane  was  rotated  so  as  to  approach  a  horizontal  direc- 
tion. (This  is  entirely  different  from  rotation  about  the  axis  OX, 
since  y  no  longer  remains  constant.) 

EXAMPLES 

1.  A  rectangular  plane  area  is  5  ft.  by  6  ft.,  the  5-ft.  side  is  horizontal,  and 
the  6-ft.  gide  vertical.     Determine  the  resultant  pressure  and  the  location  of 
the  center  of  pressure  when  :     (a)  the  top  edge  is  in  the  water  surface  ;  (6) 
the  top  edge  is  1  ft.  below  the  water  surface;  (c)  the  top  edge  is  100  ft.  below 
the  water  surface. 

Ans.  (a)  P  =  5,620  lb.,  y'  =  4  ft.;  (6)  P  =  7,500  lb.,  y'  =  4.75  ft.;  (c) 
P  =  193,000  lb.,  y'  =  103.03  ft. 

2.  Suppose  in  Fig.  20  that  we  have  a  rectangular  area  5  ft.  by  6  ft.,  that 
AB  =  6  ft.,  the  5-ft.  edge  being  normal  to  the  plane  of  the  paper,  and  that  y 

=  4  ft.     Find  the  magnitude  of  the  total  pressure  and  the  location  of  the 
center  of  pressure  when  6  has  values  of  90°,  60°,  30°,  and  10°. 

Ans.  (a)  P  =  7,488  lb.,  y'  =  4.75  ft.;  (6)  P  =  6,490  lb.;  (c)  P  =  3,744  lb.; 
(d)  P  =  1,302  lb. 

3.  Suppose  that  in  problem  (2)  y  was  variable  but  that  z  =  4  ft.     Solve 
with  values  of  6  of  90°,  60°,  30°,  and  0°. 

Ans.  (a)  P  =  7,488  lb.,  y'  =  4.75  ft.;  (6)  y'  =  5.265  ft.;  (c)  y'  =  8.375  ft.; 
(d)  ye  =  infinity,  z1  =  4  ft. 

4.  Find  the  depth  of  the  center  of  pressure  on  a  vertical  triangular  area 
whose  altitude  is  h  and  whose  base  is  b  if  :  (a)  its  vertex  lies  in  the  water 
surface;  (6)  its  base  lies  in  the  water  surface. 

Ans.  (a)  y'  =  %h;   (b)  y'  =  %h. 


22  HYDRAULICS 

23.  Lateral  Location  of  Center  of  Pressure. — For  most  prac- 
tical problems  the  depth  of  the  center  of  pressure  is  all  that  re- 
quires solution  since  the  areas  with  which  we  deal  are  usually  such 
that  a  straight  line  can  be  drawn  through  thecenters  of  all 
horizontal  lines.  In  such  cases  the  center  of  pressure  is  seen  to  lie 
on  this  line.  But  in  case  this  is  not  so  we  should  have  to  com- 
pute x'  as  in  Fig.  20,  x'  being  measured  from  any  axis  parallel  to 
trace  AB. 

Again  we  employ  moments  as  in  the  preceding  article.  If  x 
is  the  distance  of  an  element  from  the  axis  in  question  the 
moment  of  dP  is 

xdP  =  wxy  sin0  dF 
Hence  the  value  of  x'  is 

,  _    fxdP    _  w  sin  efxydF 
fdP         w  sin  BfydF 

=       -1  (11) 

yF 

This  equation  differs  from  (7)  simply  in  the  fact  that  we  have 
fxydF  instead  of  fyzdF.  The  latter  quantity  is  more  frequently 
met  with,  it  is  given  a  name,  symbolized  by  the  letter  7,  and 
values  of  7  for  different  areas  can  usually  be  found  in  tables. 
The  former  expression  is  called  "  product  of  inertia,"  is  symbolized 
by  the  letter  J,  but  owing  to  the  infrequent  use  that  is  made  of  it 
values  of  J  cannot  usually  be  obtained  save  by  integration. 
Lacking  the  knowledge  of  the  value  of  J  for  any  area,  we  should 
simply  proceed  to  evaluate  fxydF  just  as  we  should  evaluate 
fy2dF  in  case  we  did  not  know  the  value  of  7  for  the  area  in 
question. 

It  will  be  found  that  reduction  formulas  can  be  used  here  as 
with  moments  of  inertia.  If  J  indicates  the  product  of  inertia 
with  respect  to  the  intersection  of  any  two  axes,  while  a  and  6 
are  the  coordinates  of  the  center  of  gravity  of  an  area  about  which 
the  product  of  inertia  is  Jg,  it  will  be  found  that 

J  =  Jg  +  Fab. 

In  using  equation  (11)  it  must  be  noted  that  y  is  to  be  measured 
as  in  Fig.  20,  while  x  may  be  measured  from  any  axis  in  the  plane 
of  the  figure  and  perpendicular  to  OX. 


HYDROSTATIC  PRESSURE  ON  AREAS 


23 


EXAMPLES 

1.  Given  a  right  triangle  with  height,  h,  and  base,  6,  with  its  vertex  in  the 
water  surface  and  its  plane  vertical.     Find  the  value  of  y'  and  then  determine 
x':  (a)  by  inspection;  (6)  by  calculus. 

Ans.  y'  =  %h;  x'  =  %b. 

2.  Find  the  center  of  pressure  on  an  area  which  is  a  quadrant  of  a  circle. 
It  is  placed  in  a  vertical  plane  and  one  edge  lies  in  the  water  surface. 

Ans.     y  =  4r/37r;  y'  =  37rr/16;  x'  =  3r/8. 

24.  Resultant  Thrust  on  Plane  Areas. — So  far  we  have  dealt 
with  the  total  pressures  on  one  side  of  a  plane  area  alone.  Of 
course,  when  the  area  is  completely  immersed  in  a  fluid  as  shown 
in  some  of  the  previous  illustrations,  the  total  pressure  on  one 
side  is  balanced  by  that  on  the  other  and  the  net  effect  is  zero. 
But  when  the  two  sides  are  not  subjected  to  the  same  pressure, 
there  is  a  resultant  thrust  whose  value  we  desire. 


E 


So  far  we  have  considered  the  surface  of  the  liquid  as  being 
free  from  all  pressure.  Thus  in  Fig.  21  we  should  consider  the 
intensity  of  pressure  as  varying  from  zero  at  A  to  EC  at  B.  But 
in  reality  there  is  some  pressure,  in  general,  from  the  atmosphere 
acting  upon  the  water  surface  equivalent  to  a  height  of  about  34  ft. 
of  water,  and  thus  the  true  free  surface  might  really  be  at  point,  0] 
the  distance  AO  being  equal  to  the  height  of  the  water  baro- 
meter. The  absolute  intensity  of  pressure  upon  the  left-hand  side 
of  the  plane,  A  B,  therefore  varies  from  A  D  to  BE.  But  in  practical 
applications  we  desire  the  difference  between  the  pressure  on  the 
left-hand  side  and  that  on  the  right-hand  side.  But  the  pressure 
on  the  right-hand  side  is  that  due  to  the  atmosphere  and  its 
intensity  is  uniform  from  A  to  B  being  equal  to  AD'.  But  AD'  = 
AD  =  CE.  Hence  atmospheric  pressure  is  added  alike  to  both 


24  HYDRAULICS 

sides,  and  it  is  useless  to  consider  it.  Therefore,  we  neglect 
atmospheric  pressure  altogether  and  treat  the  water  surface  as  a 
true  free  surface  in  most  calculations. 

Suppose  we  have  an  area  such  as  AB  in  Fig.  22  with  a  fluid 
pressure  on  both  sides  but  of  different  intensities.  Of  course,  we ' 
could  compute  the  magnitudes  of  the  total  pressures  on  both 
sides  of  the  area  and  the  difference  would  be  the  resultant 
desired.  But  we  should  also  have  to  find  the  centers  of  pressure 
on  both  sides  and  then  locate  the  line  of  action  of  the  resultant 
of  these  two  forces.  The  following  analysis  will  indicate  a  much 
easier  solution. 

At  A  the  intensities  of  pressure  on  the  two  sides  are  AI  and 
AK.  If  IJ  be  laid  off  equal  to  AK  the  net  difference  in  the 
intensity  of  pressure  will  be  AJ.  In  similar  manner  at  B  the 
net  intensity  of  pressure  is  BF.  And  it  is  readily  seen  that, 
since  CDE  and  H KG  make  the  same  angle  with  the  vertical,  the 
values  of  HD,  AJ,  and  BF  are  equal.  Thus  the  resultant 
intensity  of  pressure  on  the  area,  AB,  is  uniform  and  equal  to  HD 
in  value.  But  HD  is  the  intensity  of  pressure  at  the  depth,  h. 
Hence  the  resultant  thrust  on  any  area  with  both  sides  com- 
pletely covered  by  the  same  liquid  is 

R  =  whF  (12) 

where  h  is  the  difference  in  level  of  the  two  liquids.  And  since 
the  net  intensity  of  pressure  is  uniform,  the  resultant  thrust  will 
act  through  the  center  of  gravity  of  the  plane  area. 

EXAMPLES 

1.  Suppose  that  a  rectangular  area  is  2  ft.  wide  by  3  ft.  high  and  that  its 
upper  edge  lies  in  a  water  surface.     What  twisting  moment  will  be  necessary 
in  a  shaft  through  A  (Fig.  21),  perpendicular  to  the  plane  of  the  paper,  to 
withstand  the  water  pressure?     It  will  be  assumed  that  the  gate  received 
no  support  save  what  the  shaft  affords,  and  that  atmospheric  pressure  acts 
alike  on  the  water  surface  and  the  right-hand  side  of  the  gate. 

Ans.  1,123  ft.-lb. 

2.  Suppose  that  the  right-hand  side  of  the  gate  in  problem  (1)  is  under  a 
vacuum  of  30  in.  of  mercury,  and  that  the  barometer  reading  is  30  in.  of 
mercury.     What  twisting  moment  would  be  required? 

Ans.  20,200  ft.-lb. 

3.  Suppose  that  the  barometer  reads  30  in.  of  mercury  and  that  the  right- 
hand  side  of  the  gate  in  problem  (1)  is  under  a  vacuum  of  20  in.  of  mercury. 
What  twisting  moment  would  be  required? 

Ans.  13,800  ft.-lb. 

4.  Suppose  the  barometer  reads  30  in.  of  mercury,  that  the  right-hand 


HYDROSTATIC  PRESSURE  ON  AREAS 


25 


side  of  the  gate  in  problem  (1)  is  under  atmospheric  pressure,  while  the  sur- 
face of  the  water  is  under  a  gage  pressure  of  50  Ib.  per  sq.  in.  What  twisting 
moment  would  be  required? 

Ans.  65,900  ft.-lb. 

6.  Suppose  in  Fig.  22  that  AB  is  a  circular  gate  of  3-ft.  diameter,  that  BC 
=  10  ft.  and  BH  =  4  ft.  Find:  (a)  magnitude  and  line  of  action  of  total 
pressure  on  left-hand  side  only;  (6)  magnitude  and  line  of  action  of  total 
pressure  on  right-hand  side  only;  (c)  resultant  thrust  on  gate. 

Ans. (a)  3,747  Ib.,  1.566  ft.  below  top  of  gate;  (6)  1,102  Ib.,  1.725  ft. 
below  top  of  gate;  (c)  2,645  Ib.,  1.500  ft.  below  top  of  gate. 

25.  Horizontal  Pressure  on  Curved  Surface. — On  any  curved 
or  irregular  area  in  general,  such  as  that  whose  trace  is  AB  in 
Fig.  23,  the  pressures  upon  different  elements  are  different  in 
direction  and  an  algebraic  or  calculus  summation  is  impossible. 


FIG.  23. 

Hence  equation  (6)  can  be  applied  only  to  a  plane  area.  But 
we  may  find  the  component  of  pressure  in  certain  directions. 
Thus  if  we  multiplied  each  dP  by  cos  6,  6  being  a  variable  angle 
which  each  elementary  force  makes  with  the  horizontal,  the  total 
horizontal  force  would  be 

px  =  fdP  cos  e.  (13) 

In  general  it  will  be  tedious  to  integrate  the  latter  and  often 
practically  impossible.  Hence  the  following  procedure  may  be 
employed 

Project  the  irregular  area  in  question  upon  a  vertical  plane, 
the  trace  of  the  latter  being  A'B'.  The  projecting  elements  are 
A  A',  BB',  etc.  It  is  seen  that  these  projecting  elements,  which  are 
all  horizontal,  enclose  a  volume  whose  ends  are  the  vertical 


26  HYDRAULICS 

plane  A  'Bf  and  the  irregular  area  whose  trace  is  A  B .  This  volume 
of  liquid  is  in  equilibrium  under  the  action  of  the  following 
forces.  Upon  the  vertical  plane  at  the  left  there  is  a  force  P', 
gravity  G'  acts  upon  the  volume  and  is  vertical,  the  pressures  on 
the  projecting  elements  are  all  normal  to  these  elements,  hence 
normal  to  P'.  Then  there  are  the  pressures  upon  the  area  in 
question  at  the  right-hand  end,  the  horizontal  component  of 
pressure  being  represented  by  Px  and  the  vertical  component  by 
Pv.  Since  we  have  a  condition  of  equilibrium  the  sum  of  all  the 
forces  in  any  direction  must  be  equal  to  zero.  But  in  a  hori- 
zontal direction  the  only  forces  are  P'  and  Px. 

Hence  Px  =  P'.  (14) 

That  is  the  horizontal  component  of  pressure  upon  any  area 
whatever  is  equal  to  the  pressure  upon  the  projection  of  the  area 
on  any  vertical  plane.  The  lines  of  action  must  also  be  the 
same. 

26.  Vertical  Pressure  on  Curved  Surface. — The  vertical  com- 
ponent of  pressure  on  an  irregular  surface  can  be  found  by  a 
method  similar  to  that  for  the  horizontal  pressure.     Thus  in 
Fig.  23  if  we  take  a  volume  of  liquid  of  which  the  area  in  question 
forms  the  base  and  vertical  elements  such  as  AD  and  BC  form 
the  sides,  we  find  the  following  forces  are  acting.     Considering 
CD  a  free  surface  the  pressure  on  the  upper  face  is  zero.     The 
pressure  on  the  lower  face  is  composed  of  the  two  components 
Px  and  Py.     Gravity,  G,  is  the   only  other  vertical  force,  the 
pressures  on  the  sides  all  being  horizontal.     Summing  up  the 
vertical  forces  and  equating  to  zero  we  have 

P,  =  O.  (15) 

Hence  the  vertical  component  of  pressure  on  any  area  whatever 
is  equal  to  the  weight  of  that  volume  of  liquid  which  would 
extend  vertically  from  the  area  to  the  free  surface. 

27.  Component  of  Pressure   in   any   Direction. — In   general 
the  component  of  pressure  in  any  direction  aside  from  horizontal 
and  vertical  cannot  be  found,  since  the  weight  of  the  volume  of 
liquid,  such  as  AA'B'B  in  Fig.  23  would  have  to  enter  the  equation. 
But  if  the  depth  of  immersion  is  great  so  that  the  pressures  on 
AB  and  A  'B'  are  great  compared  with  the  weight  Gf  the  latter 
may  be  neglected.     Hence  in  such  cases  only,  the  component  of 


HYDROSTATIC  PRESSURE  ON  AREAS  27 

pressure  in  any  direction  may  be  taken  as  the  pressure  upon  an 
area  projected  in  that  direction  upon  a  plane. 

Of  course  with  a  plane  area  the  component  of  pressure  in  any 
direction  may  be  found  by  multiplying  P  by  the  proper  function 
of  some  angle.  Or  it  may  be  convenient  to  find  it  by  the  methods 
of  Arts.  25  and  26.  Also  for  a  plane  area,  since  P  cos  0  =  (wzF) 
cos  #,  it  may  be  seen  that  the  component  of  pressure  is  the  same 
as  the  pressure  upon  an  area  of  value  F  cos  6  provided  the  center 
of  gravity  of  such  area  be  the  same  depth  as  the  center  of  gravity 
of  the  given  plane. 

28.  Resultant  Pressure  on  Curved  Surface. — In  general  there 
is  no  single  resultant  pressure  on  an  irregular  surface,  for  a  system 
of  non-parallel  and  non-coplanar  forces  does  not  usually  reduce 
to  anything  simpler  than  two  single  forces.     Thus  in  general 
Px  and  Py  are  not  in  the  same  plane  and  hence  cannot  be  com- 
bined.    But  in  some  special  cases  of  symmetrical  surfaces,  these 
two  components  will  lie  in  the  same  plane  and  hence  can  be 
combined  into  a  single  force. 

EXAMPLES 

1.  In  Fig.  24  is  shown  a  quadrant  of  a  circular  cylinder,  AB,  whose  length 
perpendicular  to  the  plane  of  the  paper  is  4  ft.  (a)  Find  the  horizontal  com- 
ponent of  pressure.  (6)  Find  the  vertical  com- 

e  /   \    T-C      i    ji  i  i  Water  Surface 

ponent  of  pressure,     (c)  Find  the  magnitude  and 
direction  of  the  resultant  water  pressure,     (d)  What     ^j= 
locates  its  line  of  action? 

29.  Pipes   under  Pressure. — If   the  in- 
ternal pressure  in  a  cylindrical  pipe  is  great 
enough  to  be  considered  in  determining  the 
thickness  of  pipe  wall  necessary,  it  will  be 
large  enough  so  that  the  weight   of   the 
water  may  be  disregarded.    Hence  according 

to  Art.  27  we  may  compute  the  resultant  pressure  in  any  direction. 
Suppose  that  in  Fig.  25,  we  pass  a  plane  XY  through  a  diameter 
of  the  pipe  as  shown.  The  total  pressure  on  one-half  of  the  pipe 
in  any  direction,  such  as  that  normal  to  XF,  will  evidently  be 
p'  X  2r  X  I,  I  being  any  length  of  pipe.  This  follows  directly 
from  Art.  27  or  may  be  seen  from  the  fact  that  the  thrust  of  the 
water  on  the  wall  of  the  pipe  normal  to  XY  must  be  balanced  by 
the  thrust  of  the  water  on  the  plane  XY.  This  pressure  will  tend 
to  rupture  the  pipe  across  the  plane  XY  and  is  resisted  by  the 


28  HYDRAULICS 

tensions  in  the  walls  of  the  pipe,  such  as  T.  Evidently  2T  = 
2p'rl.  If  the  thickness  of  the  pipe  wall  be  denoted  by  t,  and  the 
stress  induced  in  it  by  Sh,  then  T  =  Shtl.  Hence 

Sht  =  p'r  (16) 

From  (16)  the  thickness  of  wall  necessary  may  be  computed  for 
any  allowable  unit  tensile  stress.  However,  it  is  well  to  note 
that  p'  should  be  the  maximum  intensity  of  pressure  that  may 
occur  and  in  case  of  water  hammer  these  intensities  are  much 
greater  than  the  static  pressures  alone.  Also  it  may  often  be 
found  that  (16)  gives  entirely  too  thin  a  wall  to  stand  ordinary 
handling  and  to  allow  for  a  certain  amount  of  corrosion.  In 
practice  p'  is,  therefore,  increased  to  allow  for  possible  water 
hammer  and  the  thickness  determined  by  (16)  is  then  increased 


FIG.  25.  FIG.  26. 

to  a  value  necessary  for  these  other  reasons.     The  tension  in  the 
case  shown  is  called  hoop  tension. 

Referring  to  Fig.  26  it  may  be  seen  that  a  cylindrical  pipe  may 
also  be  ruptured  by  forces  parallel  to  the  axis.  Thus  the  pressure 
on  the  blank  end  is  balanced  by  the  tension  in  any  section  such  as 
XY.  The  total  pressure,  assuming  it  to  be  of  uniform  intensity, 
is  pr  X  vrr2.  And  the  tension  across  a  section  XY  is  T  = 
Si  X  2-nrt.  Hence,  equating  these  two, 

2Sit  =  p'r  (17) 

This  stress  is  called  longitudinal  tension  and  it  may  be  seen  that 
it  is  one-half  the  hoop  tension. 

For  cylinders  with  thin  walls  these  formulas  will  hold,  since 
they  assume  uniform  intensity  of  stress  across  the  metal.  But 
with  thick  walls  they  do  not  hold.  In  the  case  of  hoop  tension  in 
a  cylinder  with  thick  walls  it  is  usually  assumed  that  the  intensity 
of  stress  is  a  maximum  at  the  inner  face  and  decreases  to  zero  at 
the  outside  of  the  wall.  Also  the  elasticity  of  the  material  enters 


HYDROSTATIC  PRESSURE  ON  AREAS  29 

into  the  hypothesis.  John  Sharp1  gives  the  following  empirical 
formula  for  hoop  tension  in  a  cast-iron  cylinder  with  thick  walls 

S  loge  ~   =   P'  (18) 

where  r2  =  external  radius  and  r  =  internal  radius.  For  wrought 
iron  and  steel  cylinders  he  gives  the  empirical  expression 

2p'  (19) 

Equation  (16)  with  S  understood  as  compressive  stress  would 
also  hold  for  external  pressure  provided  the  pipe  remained  truly 
cylindrical.  But  actually  it  may  become  slightly  distorted  from 
the  cylindrical  form  and  then  there  is  a  possibility  of  sudden 
collapse.  A  large  thin  tube  which  can  stand  a  high  internal 
pressure  can  withstand  only  a  small  external  pressure.  All 
formulas  for  determining  the  strength  of  pipes  against  external 
pressure  are  purely  empirical.  So  far  no  satisfactory  expression 
has  been  deduced,  and  sufficient  data  is  lacking. 

30.  Buoyant  Force  of  the  Water  and  Flotation. — Considering 
the  body  EHDK  immersed  in  a  fluid  in  Fig.  27,  we  see  that  it  is 
acted  upon  by  gravity  and  the  pressures  from  the  surrounding 
fluid  at  least.  In  addition  there  may  be  other  forces  applied. 
On  the  upper  surface  of  the  body  the  vertical  component  of  the 
pressure,  Py,  will  be  equal  to  the  weight  of  the  volume  of  fluid 
AEHDC.  In  similar  manner  the  vertical  component  of  the 
pressure  on  the  under  surface,  P'y,  will  be  equal  to  the  weight  of 
the  volume  of  fluid  AEKDC.  It  is  evident  that  P'v  is  greater 
than  Py  and  that  the  total  vertical  force  exerted  by  the  fluid  is 
upward  and  is  equal  in  magnitude  to 

P'v  —  Py  =  weight   of    volume    AEKDC  —  weight   of    volume 

AEHDC. 

But  the  difference  between  these  two  volumes  is  the  volume  of 
the  body  EHDK.  Hence  for  any  body  immersed  in  a  fluid  such 
as  water  the  buoyant  force  of  the  water  is  equal  to  the  weight  of 
the  water  displaced. 

If  the  body  remains  in  equilibrium  in  the  position  shown  in 
Fig.  27,  when  no  other  forces  are  acting,  it  is  seen  that  G  =  P'v  — 
Py.  Hence  the  body  must  be  of  the  same  density  as  the  fluid  in 
which  it  is  immersed.  If  it  is  lighter  than  the  fluid,  a  downward 

Considerations  Regarding  Cast  Iron  and  Steel  Pipe." 


30 


HYDRAULICS 


force  will  have  to  be  applied  whose  value  is  B  —  G,  B  being  the 
buoyant  force  of  the  fluid.  If  the  body  is  denser  than  the  fluid, 
it  will  have  to  be  supported  by  a  force  whose  value  is  G  —  B. 
But  if  the  body  rests  on  the  bottom  of  a  body  of  fluid  (Fig.  28) 
in  such  a  way  that  the  fluid  does  not  have  access  to  the  under  side, 
there  will  be  no  buoyant  effect  for  then  P'v  =  zero.  Thus  in 
the  case  of  a  ship,  for  example,  sunk  in  the  mud  at  the  bottom  of  a 
body  of  water,  the  pull  T  necessary  to  raise  the  ship  is  not  only 


FIG.  27. 


FIG.  28. 


the  weight  of  the  ship  but  also  the  weight  of  the  entire  volume  of 
water  resting  on  top  of  it.     Thus  in  Fig.  28,  T  =  G  +  Py. 

If  no  external  forces  are  applied  to  a  body  which  is  lighter  than 
the  fluid;  it  will  float  on  the  surface,  such  portion  of  its  volume 
being  immersed  as  is  necessary  to  displace  an  amount  of  fluid 
equal  in  weight  to  the  weight  of  the  body. 

If  the  body  is  slightly  heavier  than  the  fluid,  it  will  sink.  If  it 
is  less  compressible  than  the  fluid  and  there  is  sufficient  depth,  it 
will  sink  until  such  a  depth  is  reached  that 
the  density  of  the  fluid  is  equal  to  its  own 
density.  If  it  is  more  compressible  than 
the  fluid  its  own  density  will  be  increased 
more  rapidly  than  that  of  the  water  and  it 
will  sink  to  the  bottom. 


EXAMPLES 


FIG.  29. 


1.  A  body  whose  volume  is  2  cu.  ft.  weighs  200  Ib. 
What  will  be  the  force  necessary  to  sustain  it  when 
it  is  immersed  in  fresh  water?     In  ocean  water? 

2.  In  Fig.  29  the  cube  A  is  12  in.  along  each  edge  and  weighs  100  Ib.     It  is 
attached  to  the  square  prism  B  which  is  6  in.  by  6  in.  by  8  ft.  and  weighs  30 
Ib.  per  cu.  ft.     What  length  of  B  will  project  above  the  water  surface? 
Ans.  1.76ft. 


HYDROSTATIC  PRESSURE  ON  AREAS 


31 


3.  A  balloon  weighs  250  Ib.  and  has  a  volume  of  10,000  cu.  ft.     When  it  is 
filled  with  hydrogen  which  weighs  0.0056  Ib.  per  cu.  ft.  what  load  will  it 
support  in  air  which  weighs  0.08  Ib.  per  cu.  ft.? 

Ans.  494  Ib. 

4.  The  specific  gravity  of  a  solid  is  0.8.     What  portion  of  its  volume  will 
be  above  the  surface  of  the  water  upon  which  it  floats? 

6.  A  body  weighs  50  Ib.  and  has  a  volume  of  4  cu.  ft.  What  vertical  force 
is  necessary  to  sink  it  beneath  the  surface  of  the  water? 

31.  Metacenter. — For  a  body  floating  on  the  surface  of  the 
water,  such  as  in  Fig.  30,  there  are  only  the  two  vertical  forces, 
its  weight  G  and  the  buoyant  force  of  the  water  B.  The  latter 
acts  through  the  center  of  gravity  of  the  water  displaced.  This 
point  is  called  the  center  of  buoyancy.  If  the  body  is  in  equilib- 
rium, these  two  forces  must  be  in  the  same  straight  line. 


FIG.  30. 

Suppose  that  by  some  external  agency  the  body  is  rolled  or  dis- 
placed through  some  angle  6.  The  center  of  gravity  is  naturally 
unchanged  in  its  position  in  the  section  but  the  center  of  buoy- 
ancy, in  general,  will  change.  Thus  G  and  M  constitute  a  couple. 
In  Fig.  30  (6)  this  is  a  righting  couple  since  it  tends  to  restore 
the  body  to  the  upright  position. 

It  may  be  seen  that  the  line  of  action  of  B  cuts  the  axis  at 
point  M.  This  point  is  called  a  metacenter.  As  the  angle  6 
varies,  the  amount  of  this  couple  will  vary  and  the  point  M  will 
also  change  its  location.  The  position  which  M  approaches  as  B 
approaches  zero  is  the  true  metacenter.  It  may  be  seen  that  if 
the  couple  is  a  righting  couple  the  point  M  must  always  be  above 
C  the  center  of  gravity.  It  is  necessary  in  ship  design  to  insure 
that  M  will  be  above  the  center  of  gravity  for  all  angles  of  heel. 
Thus  not  only  is  it  necessary  to  locate  the  true  metacenter  but 
also  to  compute  the  moment  of  the  righting  couple  for  all  values 
of  0  which  are  likely  to  be  encountered.  Further  consideration 
of  this  topic  properly  belongs  to  the  subject  of  ship  design. 


CHAPTER  IV 
APPLICATIONS  OF  HYDROSTATICS 

32.  The  Gravity  Dam. — One  of  the  most  important  of  the 
many  applications  of  hydrostatics  is  the  design  of  dams,  of  which 
there  are  several  types.  The  gravity  dam  is  one  which  depends 
for  its  stability  upon  its  weight.  A  typical  cross-section  of  such 
a  dam  is  shown  in  Fig.  31.  If  the  face  AB  is  curved  it  will  be 
necessary  to  compute  the  two  components  of  the  water  pressure, 
H  being  equal  to  the  pressure  on  a  plane  whose  trace  is  A'B 
while  V  is  the  weight  of  the  volume  of  water  represented  by 


FIG.  31. — Cross-section  of  gravity  dam. 

ABA'.  In  all  computations  it  is  customary  to  consider  a  length 
.of  dam  (perpendicular  to  the  plane  of  the  figure)  of  1  ft.  Evi- 
dently the  stability  of  a  gravity  dam  is  independent  of  the  total 
length  of  the  dam. 

The  total  water  pressure  P  combined  with  the  weight  of  the 
section  G  gives  a  resultant  pressure  on  the  base  whose  value  is 
R.  This  pressure  is  distributed  all  over  the  base  BO  but  may  be 
considered  to  have  a  single  point  of  application  C.  If  R  be 
resolved  into  two  components  at  the  point  C,  evidently  the  value 
of  the  horizontal  component  must  be  equal  to  H  while  that 

32 


APPLICATIONS  OF  HYDROSTATICS  33 

of  the  vertical  component  will  equal  G  +  V .  By  taking  moments 
of  all  the  forces  about  0  it  will  be  easy  to  locate  the  point  C. 

If  the  dam  rests  solidly  upon  impervious  rock  and  there  is  no 
leakage  of  water- along  any  plane,  or  if  a  cutoff  wall  at  B  runs 
down  deep  enough  to  stop  percolation,  and  the  base  of  the  dam 
is  well  drained,  the  above  forces  are  all  that  act  upon  the  struc- 
ture, excepting  of  course  the  support  of  the  earth  which  is  equal 
and  opposite  to  R.  But  if  water  does  have  access  to  the  under 
side  of  the  dam  there  will  be  exerted  upon  BO  a  vertical  upward 
pressure  due  to  this.  How  much  this  may  amount  to  depends 
upon  conditions.  Thus  if  water  saturates  the  foundation  but 
does  not  have  an  opportunity  of  escaping  past  0  the  whole  base 
of  the  dam  will  be  subjected  to  a  water  pressure  equal  to  BA' 
in  intensity.  But  if  the  water  can  escape  past  0  there  will  be  a 
flow  of  water  under  the  dam  and  consequently  the  pressure  must 
decrease  from  BA'  at  B  to  a  very  much  smaller  value  at  0.  It 
is  often  reasonable  to  assume  the  pressure  as  zero  at  0.  But  in 
any  event  the  admission  of  water  to  the  base  of  the  dam  tends  to 
decrease  the  safety  of  the  structure. 

It  may  be  seen  that  the  horizontal  thrust  of  the  water  H 
is  opposed  solely  by  the  friction  between  the  dam  and  the  founda- 
tion upon  which  it  rests.  If  the  coefficient  of  friction  here  be 
denoted  by  jj,  then  it  is  clear  that  if  the  dam  is  safe  against 
sliding  the  value  of  H  must  be  less  than  n(G  +  V).  The  factor, 
of  safety  against  sliding  is  the  ratio  of  the  latter  quantity  to  H. 
Any  leakage  of  water  under  the  base  of  the  dam  decreases  the 
pressure  between  the  dam  and  the  material  upon  which  it  rests 
and  thus  tends  to  decrease  the  frictional  resistance.  The 
frictional  resistance  can  be  increased  by  sinking  portions  of  the 
dam  into  trenches  such  as  in  the  case  of  the  cutoff  wall  at  B. 

If  it  were  possible  for  the  dam  to  act  as  a  rigid  body  under  all 
circumstances  it  could  then  fail  by  overturning  about  0  as  an  axis. 
It  is  seen  that  with  0  as  a  center  of  moments,  H  tends  to  overturn 
the  dam  but  is  resisted  by  G  and  V.  If  water  pressure  acted  upon 
the  base  it  would  also  tend  to  overturn  the  dam.  The  factor 
of  safety  against  overturning  is  the  ratio  of  the  moment  of  G  +  V 
to  the  moment  of  H  and  the  water  pressure  on  the  base,  if  any  is 
allowed  for.  However,  before  a  masonry  dam  of  any  size  would 
overturn,  the  material  along  the  base  near  O  would  be  crushed 
due  to  the  high  intensity  of  pressure  it  would  be  under.  Thus 
although  the  point  C  might  be  to  the  left  of  0  in  Fig.  31  so  that 


34 


HYDRAULICS 


the  structure  is  safe  against  overturning,  the  base  would  still 
not  be  safe  against  crushing.  Hence,  the  second  consideration 
of  the  stability  of  the  dam  is  not  as  to  whether  it  will  or  will 
not  overturn  but  is  concerning  the  distribution  of  stresses  along 
the  base  BO. 


B         A 


(O 


Referring  to  Fig.  32,  a  uniform  intensity  of  stress  p'  distributed 
over  an  area  represented  by  AB  gives  a  resultant  pressure  P  applied 
midway  between  A  and  B.  If,  however,  the  stress  varies  uni- 
formly from  P'A  at  A  to  zero  at  B,  the  resultant  P  will  pass  through 
a  point  one-third  the  distance  from  A  to  B.  If  the  total  pressure 


From  a  photograph  by  the  author. 
FIG.  33. — Concrete  dam  at  Crystal  Springs  Lake,  California.     145  feet  high. 

P  has  the  same  value  in*  both  cases,  it  is  clear  that  the  intensity 
of  pressure  at  A  is  greater  in  the  latter  case  than  in  the  former. 
And  if  P  is  applied  at  a  point  less  than  one-third  the  distance 
from  A  to  B,  the  intensity  of  stress  will  be  still  greater  at  A  and 
at  B  the  intensity  of  stress  p's  will  be  opposite  in  sign  to  that  at 


- 
APPLICATIONS  OF  HYDROSTATICS  35 

A.  It  is  thus  clear  that  it  is  desirable  to  have  the  resultant  pres- 
sure pass  as  nearly  through  the  midpoint  as  possible.  And  if 
tensile  stresses  are  to  be  avoided  the  resultant  pressure  must  be 
kept  within  the  middle  third.  As  masonry  is  not  supposed  to  en- 
dure tensile  stresses,  it  is  customary  to  so  design  the  dam  that  the 
resultant  pressure  falls  within  the  middle  third  of  any  section. 

It  is  not  only  necessary  to  undertake  such  an  analysis  of  the 
dam  as  a  whole  but  also  to  investigate  the  stability  of  all  portions 
of  the  dam  with  respect  to  any  horizontal  plane.  In  all  such 
studies  the  maximum  height  of  water  should  be  assumed.  But 
also  the  pressures  should  be  determined  when  the  reservoir  is 
empty  as  the  inner  face  of  the  dam  might  then  be  subjected  to 
excessive  vertical  stresses. 

33.  The  Framed  Dam. — Contrasted  with  the  gravity  dam  we 
have  the  framed  dam  shown  in  Fig.  34  which  depends  for  its 


FIG.  34. — Framed  dam. 

stability  upon  the  strength  of  its  members.  It  consists  of  a 
water-tight  deck  AB  supported  by  struts,  trusswork,  or  buttresses 
at  certain  intervals  along  the  length  of  the  dam  (perpendicular 
to  the  plane  of  the  figure).  The  deck  is  always  inclined  so  that 
the  weight  of  the  water  upon  it  may  hold  the  structure  down  and 
increase  the  factor  of  safety  against  sliding. 

34.  The  Arch  Dam. — In  the  case  of  a  short  high  dam  in  a 
situation  where  firm  support  can  be  had  from  the  walls  on  either 
side  the  arch  dam  is  desirable.  It  is  designed  to  withstand  the 
water  pressure  by  pure  arch  action  and  to  transmit  the  pressures 
to  the  abutments  at  either  end.  The  material  in  an  arch  dam  is 
usually  much  less  than  in  a  pure  gravity  dam  but  any  arch  dam 


36 


HYDRAULICS 


acts  to  some  extent  as  a  gravity  dam.     Its  analysis  is  not  within 
the  scope  of  this  text. 


From  a  photograph  by  the  author. 

FIG.  35. — Lake  Spau.ld.ing,  Cal.,  variable  radius  arch  dam.     Ultimate  height 

will  be  325  ft. 


JS^^I^     ^00*°        '      $ '     /effect  becomes  imper-V       \,    \  %..  Xx 
^^^  ^       **       /       «T    /  vious  lacustrine Jclay\        \      \     *>S        ^. 


"West  Side      Gate  Tower 


Approximate  Line  of  Bed  Rock  2& 

East  Side 


Spillway 


Ground  Line  - 

Concrete  Culvert 


FIG.  36. — Section  of  Calaveras  earth  dam. 

35.  The    Earth    Dam. — Under   favorable    circumstances    the 
earth  dam  is  a  very  economical  type.     A  typical  section  of  such 


APPLICATIONS  OF  HYDROSTATICS 


37 


a  dam  may  be  seen  in  Fig.  36.     The  slopes  on  both  the  upstream 
and  downstream  faces  are  less  than  the  angle  of  repose  of  the 


From  a  photograph  by  the  author. 

FIG.  37. — Upstream  face  of  San  Andreas  earth  dam.     90  ft.  high. 


From  a  photograph  by  the  author. 

FIG.  38. — Incompleted  Calaveras  earth  dam.     Ultimate  crest  will  be  at 
dotted  line  making  it  the  highest  earth  dam  in  the  world. 

material  used.  In  order  to  make  such  a  dam  water-tight  it  is 
provided  with  an  impermeable  core  which  may  be  a  thin  vertical 
wall  of  concrete  or  other  material,  or,  as  in  Fig.  36,  it  may  be 


38  HYDRAULICS 

obtained  by  depositing  fine  earth  under  water.  Fig.  38  shows  the 
pool  of  water  in  the  center  of  the  dam  where  such  a  core  is  being 
formed.  There  is  little  mathematical  analysis  to  be  made  for 
such  a  dam.  The  main  problems  are  those  of  construction  and 
careful  selection  of  the  materials  employed. 

36.  Additional  Notes  on  Dams. — In  most  cases  there  are 
times  when  there  is  an  excessive  quantity  of  water  that  must  be 
disposed  of,  usually  by  allowing  it  to  flow  over  a  spillway  that 
is  provided  for  that  purpose.  The  spillway  may  be  located  at  a 
different  place  from  the  dam  so  that  no  water  ever  overtops  the 
latter  as  will  be  the  case  in  Fig.  35.  Again  the  spillway  may 


From  a  photograph  by  the  author. 

FIG.  39. — Low  dam  at  Ithaca,  N.  Y. 

occupy  a  portion  of  the  crest  of  the  dam  as  in  Fig.  33  where  the 
spillway  can  be  seen  in  the  middle.  In  other  cases,  such  as  in 
Figs.  36,  37,  and  38,  the  spillways  are  located  at  one  end  of  the 
dam  and  consist  of  rectangular  canals  through  which  the  flood 
waters  are  discharged.  But  in  Fig.  39  it  may  be  seen  that  the 
entire  crest  of  the  dam  is  used  for  a  spillway.  This  dam  also 
shows  the  curved  face  that  is  provided  to  minimize  the  scouring 
effect  of  the  waterfall  upon  the  bed  of  the  stream  at  the  toe  of  the 
dam.  For  it  must  be  recognized  that  water  in  falling  over  a  dam 
acquires  kinetic  energy  that  must  be  expended  in  some  way  and 
unless  suitable  provision  is  made  for  this  it  may  be  expended  in 
undermining  the  dam  itself. 


APPLICATIONS  OF  HYDROSTATICS 


39 


37.  Flashboards. — In  storing  water  by  means  of  a  dam  it  is 
desirable  to  keep  the  water  level  as  high  as  possible  without 
flooding  any  lands  upstream.  If,  therefore,  the  crest  of  the  dam 
were  located  at  the  elevation  allowable  under  normal  conditions 
it  would  be  excessively  high  in  times  of  flood.  In  order  to  over- 
come this  difficulty  movable  devices  are  employed  called  flash- 
boards,  movable  crests,  and  various  other  names  (Fig.  40). 
These  are  all  schemes  for  increasing  the  height  of  the  dam  by 
equipment  which  can  be  removed  when  necessary.  In  some  cases 
they  work  automatically,  being  either  washed  away  when  the 
water  reaches  a  certain  stage  or  caused  to  drop  to  a  horizontal 
position.  Other  types  require  removal  by  hand  in  such  emer- 
gencies. After  the  flood  is  past  and  the  drier  season  comes  on 


T 


\ 


\ 


FIG.  40. — Flashboard. 

they  may  be  replaced  again.  Some  of  these  are  entirely  auto- 
matic in  their  action  as  in  the  case  of  the  Stickney  automatic  crest 
outlined  in  Fig.  41.  We  have  here  two  planes  AB  and  BC 
rigidly  connected  and  rotating  about  B.  The  water  pressure  on 
AB  together  with  the  weight  of  the  shutters  and  the  additional 
weight  added  at  C  tend  to  rotate  the  device  in  one  direction  but 
that  is  opposed  by  the  pressure  of  the  water  on  BC.  By  a 
suitable  adjustment  of  area  and  weights  it  is  possible  to  keep  this 
crest  in  the  position  shown  until  the  water  reaches  the  level  of  A. 
Then  the  pressure  on  AB  may  be  sufficient  to  cause  it  to  drop  to 
the  position  A'BC'.  Hence  the  crest  of  the  dam  will  then  be 
reduced  to  the  height  of  B}  and  the  flood  water  will  pour  over  the 
shutter  BA'  and  hold  it  down.  But  when  the  excess  waters  have 
passed  and  the  water  level  drops  to  B,  or  thereabouts,  the  pres- 


40 


HYDRAULICS 


sure  on  BC',  no  longer  opposed  by  that  on  BA',  will  raise  the 
crest  to  the  initial  position. 

A 


FIG.  41. — Automatic  dam  crest. 


38.  PROBLEMS 

1.  The  intake  tower  in.  Fig.  42  will  be  surrounded  by  water  when  the 
reservoir   is   filled  and  the  outflow  of  water  will  take  place  through  the 
openings  provided  in  the  tower.     Assume  one  of  these  gates  to  be  3  ft.  wide 
and  4  ft.  high  and  to  weigh  1,000  Ib.     When  the  inside  of  the  tower  is  sub- 
jected to  the  pressure  of  the  air  only,  what  vertical  pull  on  a  gate  rod  will  be 
necessary  to  open  the  gate  when  the  water  stands  10  ft.  above  its  top,  if 
the  coefficient  of  friction  between  the  gate  and  its  guides  is  0.3? 

2.  The  valve  in  Fig.  43  is  34  in.  in  diameter.     If  it  is  closed  and  under  a 
pressure  of  1,000  ft.  of  water  on  one  side  and  atmospheric  pressure  on  the 
other,  what  pull  will  have  to  be  exerted  on  the  valve  stem  to  open  it  if  the 
coefficient  of  friction  is  0.4? 


APPLICATIONS  OF  HYDROSTATICS 


41 


From  a  photograph  by  F.  H.  Fowler. 

FIG.  42. — Intake  tower  at  Elizabeth  Lake  Reservoir  on  Los  Angeles  Aqueduct. 


From  a  photograph  by  the  author. 

FIG.  43. — A  34-in.  high-pressure  gate  valve  in  the  shop  of  the 
Pelton  Water  Wheel  Co. 


42 


HYDRAULICS 


3.  Find  the  magnitude  and  point  of  application  of  the  resultant  pressure 
on  the  2-ft.  circular  gate  shown  in  Fig.  44. 

4.  The  gate  AB  in  Fig.  45  rotates  about  an  axis  through  B.     If  the  width 
is  4  ft.,  what  torque  applied  to  the  shaft  through  B  is  required  to  keep  the 
gate  shut? 

6.  What  value  of  b  in  Fig.  46  is  necessary  to  keep  the  masonry  wall  from 
sliding?  Masonry  weighs  150  Ib.  per  cu.  ft.  and  the  coefficient  of  friction 
equals  0.4.  Will  it  also  be  safe  from  overturning?  If  it  has  a  factor  of 
safety  against  sliding  of  2,  where  will  the  resultant  of  the  water  pressure 
and  its  weight  cut  the  base? 


FIG.  44. 


i 


FIG.  46. 


6.  In  the  framed  dam  shown  in  Fig.  47,  the  struts  CD  are  placed  5  ft. 
apart  along  the  dam  (perpendicular  to  the  plane  of  the  figure).     What  will 
be  the  load  on  each  strut?     What  will  be  the  value  of  the  reaction  at  A? 
If  the  length  BE  is  4  ft.  and  the  depth  of  the  water  flowing  over  the  crest  at 
E  is  3  ft.  what  will  be  the  load  on  the  strut? 

7.  Assume  the  weight  of  the  dam  in  Fig.  48  to  be  150  Ib.  per  cu.  ft.,  that 
there  is  no  seepage  of  water  under  its  base,  and  that  the  coefficient  of  friction 
between  the  dam  and  the  material  upon  which  it  rests  is  0  6.     For  1  ft. 


T 


FIG.  47. 


length  compute :  (o)  Horizontal  component  of  water  pressure,  (b)  Vertical 
component  of  water  pressure,  (c)  Weight  of  dam.  (d)  Is  it  safe  against  slid- 
ing? (e)  Is  it  safe  against  overturning?  (/)  Where  does  the  resultant  of 
the  water  pressure  and  the  weight  of  the  dam  cut  the  base? 

8.  In  Fig.  40  the  flashboard  AB  rests  against  a  solid  block  at  B  but  there 
is  a  pin  at  either  end  at  A  which  is  breakable.  If  the  length  of  a  section  of 
flashboard  is  6  ft.,  what  must  be  the  shearing  strength  of  the  pins  if  they  give 
way  when  the  water  level  reaches  A  ? 


APPLICATIONS  OF  HYDROSTATICS 


43 


9.  In  Fig.  41  what  weight  must  be  added  at  C  per  foot  of  length  in  order 
that  the  crest  may  drop  when  the  water  level  reaches  A  ?     Neglect  the  weight 

.of  the  rest  of  the  movable  crest,  and  assume  BC  =  7.5  ft. 

10.  Figure  49  shows  a  cylindrical  tank.     What  is  the  total  pressure  on  the 
bottom  ?     What  is  the  total  pressure  on  the  annular  surface  A- A  ?     Find  the 
maximum  intensity  of  longitudinal  tensile  stress  in  side  walls  B-B. 


U-12Diam.->| 

I 


T 

12" 


12" 

I 


K 24Diam. H 

FIG.  49. 

11.  A  pipe  line  3  ft.  in  diameter  is  to  carry  water  under  a  pressure  of 
1,000  ft.     If  the  allowable  tensile  stress  is  20,000  Ib.  per  sq.  in.,  what  should 
be  the  thickness  of  steel  used? 

12.  With  the  thickness  of  metal  computed  in  the  preceding  problem  what 
would  be  the  tensile  stress  across  a  circumferential  section  if  a  valve  was 
closed,  the  pressure  on  the  other  side  of  it  being  atmospheric? 


CHAPTER  V 


HYDROKINETICS 

39.  Actual  and  Ideal  Conditions. — From  the  standpoint  of 
pure  mechanics  the  subject  of  hydrokinetics  is  rather  unsatis- 
factory. This  is  due  to  the  fact-  that  so  many  assumptions  are 
necessary,  many  of  which  are  known  not  to  be  true.  Thus  in 
the  greater  portion  of  the  work  all  particles  of  water  in  any  cross- 
section  of  a  flowing  stream  are  assumed  to  move  in  parallel 
paths  and  with  equal  velocities.  This  is  shown  in  Fig.  50,  a 


0 

^^^   =^ 

x      1 

'>\B     \ 

y 

1          I 

0'                C 

FIG.  51. 

particle  of  water  at  point  0  moving  along  the  axis  of  a  pipe  with 
a  velocity  OB.  But  every  other  particle  of  water  across  section 
0'00f  is  assumed  to  move  with  the  same  velocity  giving  us  the 
velocity  curve  ABC,  in  this  case  a  straight  line.  But  it  is  well 
known  that  in  a  pipe  the  actual  velocity  curve  is  similar  to  ABC 
in  Fig.  51,  the  velocity  of  a  particle  of  water  at  0  in  the  center 

//ffffffffffffffffffffffffffffffff^fffffffffff^f^7       °f  ^ne  PiPe  being  OB  while 

^ — ^^^  ^     that  of  a  particle   near    the 

wall  of  the  pipe  is  O'A.  Ex- 
periment shows  that  in  gen- 
eral OB  is  about  twice  the 
value  of  O'A  and  that  the 
mean  velocity  of  all  the  particles  is  about  0.84  OB.  It  is  this 
mean  velocity  that  is  actually  used  in  our  computations.  Hence 
our  results  are  based  upon  a  velocity  which  is  possessed  by  only  a 
few  of  the  particles  of  water,  the  greater  portion  of  them  moving 
with  either  higher  or  lower  velocities.  We  usually  do  not  attempt 
to  deal  with  the  actual  velocity  curve  ABC  of  Fig.  51  because  we 

44 


FIG.  52. 


HYDROKINETICS 


45 


have  no  assurance  as  to  its  exact  nature  in  every  case  and,  if  we 
did,  our  equations  would  be  too  complicated  for  practical  use. 

But  in  Fig.  51  all  particles  of  water  have  been  assumed  to  be 
moving  in  straight  lines  parallel  to  the  axis  of  the  pipe,  which  we 
know  is  very  seldom  the  case.  In  fact  the  path  of  a  given  par- 
ticle is  very  irregular  as  is  shown  in  Fig.  52  and  at  the  instant 


From  a  photograph  by  the  author. 

FIG.  53.  Showing  vortices  on  surface  of  canal. 

in  question  a  particle  at  point  0  may  be  moving  with  some 
velocity  OD.  But  in  most  practical  problems  we  are  concerned 
with  OB  which  is  the  axial  component  of  the  true  velocity.  Thus 
not  only  do  our  equations  ordinarily  deal  with  a  mean  velocity, 
but  they  deal  with  a  component  of  the  true  velocity.  Instead 
of  water  flowing  in  parallel  threads  the  true  phenomena  has  been 


46  HYDRAULICS 

very  aptly  compared  to  the  motion  of  a  cloud  of  feathers  blown 
along  by  the  wind.  Water  tends  to  travel  in  vortices  as  may  often 
be  observed  upon  the  surface  of  an  open  stream  such  as  the  canal 
shown  in  Fig.  53.  In  this  particular  scene  the  water  was  flow- 
ing with  a  moderate  velocity  (about  3  miles  an  hour)  over  a 
reasonably  smooth  bed  but  the  surface  was  covered  with  little 
vortices. 

Since  actual  conditions  depart  so  widely  from  the  ideal  con- 
ditions assumed  by  our  imperfect  theory  we  can  expect  our 
theory  to  provide  little  more  than  a  framework  upon  which  may 
be  hung  the  results  of  experimental  investigation. 

The  mean  velocity  at  any  section  (strictly  the  mean  axial 
component  of  velocity)  is  obtained  by  dividing  the  total  rate  of 
discharge  by  the  total  area  of  the  section.  That  is  V  =  q/F. 

EXAMPLES 

1.  Experiment  indicates  that  the  velocity  curve  ABC  of  Fig.  51  is  approxi- 
mately a  semi-ellipse  and  that  OB  is  about  twice  O'A.     Assuming  this  to 
be  so,  find  the  ratio  between  the  mean  velocity  and  the  maximum  velocity. 
(The  total  rate  of  discharge  is  f  VdF  and  the  value  of  this  integral  is  the 
volume  of  the  solid  O'A  SCO'.     Dividing  the  solid  by  the  area  of  the  base, 
irr2,  we  should  have  the  mean  ordinate  or  in  this  case  the  mean  velocity. 
The  volume  of  an  ellipsoid  is  two-thirds  that  of  the  circumscribing  cylinder.) 

Ans.  0.833. 

2.  A  stream  is  divided  into  five  equal  areas  and  the  mean  velocity  of 
each  portion  is  found  by  some  method.     These  velocities  are  3,  3,  4,  4,  and 
5  ft.  per  sec.     What  is  the  mean  velocity  of  the  entire  stream? 

Ans.  3.80  ft.  per  sec. 

3.  Suppose  that  the  areas  are  not  equal  but  have  values  of  2.5,  2.5,  2.0, 
2.0,  and  1.0  sq.  ft.  while  the  velocities  are  3,  3,  4,  4,  and  5  ft.  per  sec.,  respect- 
ively.    What  is  the  total  rate  of  discharge?     What  is  the  mean  velocity? 

Ans.  36  cu.  ft.  per  sec.;  3.60  ft.  per  sec. 

40.  Critical  Velocity. — The  path  followed  or  assumed  to  be 
followed  by  a  single  particle  of  fluid  is  called  a  stream  line.  It 
has  been  found  that  for  very  low  velocities  the  stream  lines  are 
straight  parallel  lines  as  shown  in  Figs.  50  and  51,  but  that  as 
soon  as  a  certain  velocity  is  exceeded  the  flow  becomes  turbulent 
or  sinuous  as  in  Fig.  52.  The  velocity  at  which  the  change 
occurs  is  called  the  critical  velocity.  The  value  of  the  critical 
velocity  is  affected  by  the  temperature  and  also  the  size  of  the 
tube  or  pipe ;  the  larger  the  latter  the  lower  the  critical  velocity. 
For  ordinary  size  pipes  with  which  the  engineer  has  to  deal  the 
critical  velocity  is  so  low  that  its  value  is  of  no  interest, 


HYDROKINETICS 


47 


41.  Steady  Flow. — By  steady  flow  is  meant  that  at  any  point 
in  a  stream  all  conditions  remain  constant  with  respect  to  time. 
This  does  not  mean  that  the  conditions  at  any  one  point  are 
necessarily  like  those  at  some  other  point. 

Unsteady  flow  is  met  with  in  cases  where  change  is  taking 
place.  Thus  suppose  a  pipe  line  is  flowing  full  of  water  and  a 


From  a  photograph  by  the  author. 

FIG.  54.  The  Los  Angeles  Aqueduct. 

valve  is  closed  suddenly  at  its  lower  end.  The  velocity  of  the 
water  would  be  brought  to  zero  and  in  so  doing  there  would  be 
certain  pulsations  of  pressure,  which  if  violent  enough  would  be 
recognized  as  water  hammer.  While  such  changes  are  in  progress 
we  should  have  unsteady  flow.  Again  suppose  that  a  gate  is 
opened  so  as  to  admit  water  into  an  open  canal  originally  empty. 
As  the  canal  filled  with  water  the  level  at  any  point  would  stead- 


48  HYDRAULICS 

ily  rise  and  also  the  velocity  would  in  general  be  changing  at  all 
points.  While  such  changes  were  under  way  the  flow  would  be 
unsteady.  But  when  equilibrium  is  finally  established,  the  water 
level  at  any  point  and  the  velocity  of  flow  across  any  section  no 
longer  vary  from  time  to  time  and  we  then  have  steady  flow. 

In  the  strictest  sense  of  the  word  steady  flow  is  seldom  met 
with  in  ordinary  engineering  work  as  it  would  be  found  only 
with  velocities  below  the  critical  velocity.  For  with  all  veloci- 
ties above  the  critical  we  have  continual  fluctuations  of  flow 
at  any  point  due  to  the  irregular  motion  of  the  individual  par- 
ticles. It  is  for  this  reason  that  manometers  or  pressure  gages 
attached  to  pipes,  in  which  water  is  flowing,  continually  pulsate. 
Another  evidence  may  be  seen  in  Fig.  54,  the  dark  band  on  either 
side  of  the  water  being  where  the  latter  has  wet  the  concrete  by 
wave  action. 

For  all  practical  purposes  we  disregard  these  slight  fluctuations 
at  individual  points.  If  the  average  conditions  over  the  entire 
section  are  reasonably  constant  with  respect  to  time,  we  consider 
the  flow  as  steady.  While  problems  of  unsteady  flow  are  often 
problems  of  great  practical  value,  especially  in  connection  with 
the  speed  regulation  of  water  power  plants,  they  are  rather 
difficult  of  mathematical  treatment.  Fortunately  they  are  not 
as  common  as  the  more  simple  problems  of  steady  flow.  For  the 
most  part  this  text  will  be  devoted  to  the  latter. 

42.  Rate  of  Discharge. — The  volume  of  water  flowing  across 
any  section  per  unit  time  is  called  the  rate  of  discharge.     It  must 
not  be  confused  with  velocity,  since  it  is  the  product  of  the  cross- 
section  area  of  the  stream  and  the  velocity  of  flow  across  the 
section.     It  may  be  expressed  in  various  units  such  as  cubic 
feet  per  minute,  gallons  per  day,  etc.,  depending  upon  the  custom 
in  that  particular  class  of  work.     In  the  foot-pound  second  sys- 
tem of  units  such  as  are  employed  in  this  text  it  would  naturally 
be  in  cubic  feet  per  second.     This  is  often  called  " second  foot" 
for  brevity  and  written  as  "sec.  ft."1 

43.  Equation  of  Continuity. — In  Fig.  56  it  is  apparent  that  the 
volume  of  water  between  any  two  sections  such  as  (1)  and  (2) 
must  remain  constant  if  the  flow  is  steady.     Hence  it  follows 
that  the  rate  at  which  water  flows  in  at  (1)  must  be  equal  to  the 
rate  at  which  it  flows  out  at  (2),  otherwise  there  would  be  a  change 

1  In  irrigation  work  in  India  the  term  "cusec"  has  gained  acceptance  for 
this  rate  of  discharge. 


HYDROKINETICS  49 

in  the  volume  contained  between  the  two  sections.  Thus  we 
may  say  that  for  steady  flow,  qi  =  q%. 

If  the  flow  is  unsteady  this  is  not  necessarily  so.  For  suppose 
that  the  closure  of  a  gate  above  (1)  shut  off  the  flow  of  water  at 
(1),  we  would  still  find  water  flowing  for  a  time  past  (2)  though 
at  the  expense  of  the  volume  stored  between  the  two  sections. 
Hence  in  case  of  unsteady  flow,  where  the  volume  in  any  distance 
is  changing,  the  equation  of  continuity  no  longer  applies. 

The  equation  of  continuity  states  that  for  steady  flow 

q  =  FiVi  =  F2VZ  = =  FV  =  constant      (20) 

This  equation  justifies  the  use  of  the  term  "  rate  of  discharge  "  for 
the  rate  of  flow  across  any  section  even  though  it  be  in  the  middle 
of  a  length  of  pipe  or  at  some  point  in  a  river.  For  at  some 
ultimate  point  the  pipe  or  stream  actually  discharges  in  the  usual 
sense  of  the  word.  And  the  rate  of  discharge  at  this  point  is 
equal,  if  the  flow  be  steady,  to  the  rate  of  volume  flow  at  all 
sections  throughout  the  stream. 

EXAMPLES 

1.  In  Fig.  55  the  portion  of  pipe  between  A  and  C  is  the  frustum  of  a 
right  circular  cone  with  vertex  at  0.  If  the  rate  of  discharge  is  10  cu.  ft. 
per  sec.,  what  are  the  velocities  at  A,  B,  and  C?  Between  A  and  C  the 
velocity  will  vary  as  what  function  of  the  distance  from  0?  What  kind  of  a 
curve  would  be  necessary  between  A  and  C  if  the  velocity  were  to  decrease 
uniformly  with  respect  to  distance? 


2.  The  canal  shown  in  Fig.  53  is  14.5  ft.  wide  and  4.2  ft.  deep.     If  the 
velocity  of  the  water  is  3.5  miles  per  hour,  what  will  be  the  rate  of  dis- 
charge in  cubic  feet  per  second? 

3.  The  water  in  the  canal  of  problem  (2)  finally  flows  down  a  steel  pen- 
stock (Fig.  Ill)  which  is  52  in.  in  diameter.     What  is  the  velocity  of  flow? 

4.  At  the  end  of  the  pipe  line  in  problem  (3)  the  water  is  discharged 
through  four  nozzles  the  jets  from  which  are  approximately  7  in.  in  diameter. 
What  is  the  jet  velocity? 

44.  General  Equation  for  Steady  Flow. — In  the  case  of  steady 
flow  we  may  derive  a  very  useful  equation  commonly  known  as 


50 


HYDRAULICS 


Bernoulli's  theorem  in  honor  of  Daniel  Bernoulli  who  proposed 
it  in  1738.  Referring  to  Fig.  56  let  us  consider  the  motion  of  a 
small  volume  of  water  whose  cross-section  area  is  dF  and  whose 
length  is  dl.  We  shall  apply  the  principle  that  the  net  work 
done  by  all  the  forces  is  equal  to  the  change  in  kinetic  energy. 
Let  dl  be  the  distance  moved  in  some  time  dt  during  which 
time  the  pressure,  elevation,  and  velocity  change  by  amounts  dp', 
dz,  and  dV,  respectively.  The  weight  of  this  elementary  volume 
is  wdFdl  and  its  kinetic  energy  changes  from  (wdFdl)  V2/2g 
to  (wdFdl)(V  +  dV)2/2g.  Expanding  (7  +  dV)2  we  have 


FIG.  56. 


V2  +  2V dV  +  dV2.  Since  dV  is  an  infinitesimal  quantity 
the  last  term  dV2  is  negligible  in  comparison  with  other  terms. 
Hence  the  change  in  kinetic  energy  equals 


wdFdl(V2  +  2VdV)/2g  -  wdFdlV2/2g 


or 


A  K.E.  = 


wdFdl 


2VdV. 


The  pressures  on  the  two  ends  do  work  the  values  of  which 
are  p'dFdl  and  —  (pr  +  dp')dFdlt  or  a  net  amount  of  work 
of  —  dp'dFdl.  Replacing  p'  by  its  equivalent  wp  we  have  the 
net  work  done  by  the  pressures  on  the  ends  to  be  —  wdFdldp. 
The  work  done  by  gravity  is  evidently  —  (wdFdl)  dz,  the  minus 
sign  being  used  since  dz  is  negative,  indicating  a  decreasing  value 
of  z  as  shown  in  Fig.  56. 


HYDROKINETICS  51 

Due  to  viscosity  there  is  also  work  done  by  virtue  of  the  re- 
sistance offered  to  flow  by  adjacent  particles  of  water.  The  exact 
nature  of  this  phenomenon  defies  accurate  mathematical  ex- 
pression but  it  seems  reasonable  to  assume  that  it  is  proportional 
to  the  size  (and  hence  weight)  of  the  volume  of  water  considered. 
Thus  let  k"  be  any  arbitrary  factor  such  that  k"(wdFdl)  rep- 
resents the  resistance  opposing  the  motion  of  the  elementary 
volume  in  Fig.  56.  Then  the  work  done  by  this  frictional  re- 
sistance will  be  -  k"(wdFdl)dl. 

Now  the  net  work  done  by  all  the  forces  acting  on  the  ele- 
mentary volume  is 

-(wdFdl)dp  -  (wdFdl)dz  -  k"(wdFdl)dl. 

Equating  the  net  work  done  to  the  change  in  kinetic  energy  we 
have 

-  (wdFdl)dp  -  (wdFdl)dz  -  k"(wdFdl)dl  =  ^^VdV. 

y 
Factoring  out  the  common  term  (wdFdl)  we  may  write 

/-P.          /•«  ri        i  rv* 

-  I     dp  -       dz  -  k"  I   dl  =  -  I     VdV. 

Jpi  Jzi  Jo  Q  JVi 

Integrating  between  limits  and  rearranging  we  have 

Fi2  F22 

Pi  +  zi  +  -^  -  k"l  =  p2  +  z2  +  -^- 

Bernoulli's  theorem  is  really  this  equation  with  the  term  k"l 
omitted  but,  since  all  real  fluids  are  viscous,  it  is  impossible 
for  flow  to  take  place  without  fluid  friction  and  hence  such  a 
term  should  always  be  included. 

The  friction  term  is  seen  to  be  proportional  to  the  distance 
moved  through,  as  we  should  expect.  Experiment  indicates 
that  it  is  also  some  function  of  the  velocity  and  we  shall  there- 
fore insert  such  a  term,  realizing  that  it  is  really  an  empirical 
expression.  Thus  instead  of  k"l  we  shall  have  k'lV*.  For 
convenience  later  we  shall  introduce  the  term  2g  into  the  expres- 
sion and  let  k  =  k'l2g  so  that  the  friction  loss  may  be  given  as 
kVn/2g.  Hence  we  may  write  the  general  equation  as 

Fi 2  Vn  Fo2 

*+*+-£ -*--.„  *+*+-£.  (21) 

In  case  the  velocity  varies  between  (1)  and  (2)  the  V  for  the 
friction  term  might  be  taken  as  the  average  velocity,  or  by  using 


52  HYDRAULICS 

a  suitable  value  of  k  it  may  be  written  as  Fi  or  F2.  In  practical 
work  the  difficulty  of  using  equation  (21)  lies  largely  in  estimat- 
ing proper  values  of  k  and  n,  and  it  is  necessary  to  rely  entirely 
upon  experimental  evidence. 

45.  Use  of  the  Word  "Head." — Examining  each  term  of 
equation  (21)  in  detail  we  find:  The  term  p  indicates  intensity 
of  pressure  expressed  in  feet  of  water,  hence  it  is  a  linear  quantity 
and  indicates  the  height  of  a  column  of  water  necessary  to  pro- 
duce the  given  pressure.  There  may  be  no  such  real  height  of 
water  in  the  problem,  as  in  the  case  of  a  small  volume  of  water 
enclosed  within  a  cylinder  and  subjected  to  pressure  by  a  piston. 
The  quantity  p  is  called  pressure  head. 

The  elevation  of  a  point  above  any  arbitrary  datum  plane 
is  indicated  by  z.  It  is  a  linear  quantity  and  in  our  system  of 
units  it  should  be  expressed  in  feet.  It  is  called  elevation  head 
or  potential  head. 

The  third  term  V2/2g  may  also  be  seen  to  reduce  to  a  linear 
quantity  when  we  analyze  the  units  involved  in  V  and  g.  The 
.linear  quantity  equivalent  to  V2/2g  is  the  height  through  which 
a  body  might  fall  in  a  vacuum  from  rest  and  acquire  the  velocity 
V.  In  many  cases  it  is  a  purely  artificial  quantity  in  that  there 
is  no  actual  height  in  the  figure  illustrating  the  problem  that 
gives  any  indication  of  its  value.  It  may  be  called  velocity  head. 

Since  all  the  other  quantities  in  equation  (21)  are  in  linear 
•dimensions,  or  feet  in  our  system  of  units,  it  follows  that  kVn/2g 
must  also  be  in  feet.  It  may  be  called  the  lost  head,  and  is 
represented  by  the  letter  H'. 

The  sum  of  the  pressure,  elevation,  and  velocity  heads  at  any 
section  is  called  the  total  or  the  effective  head  at  that  section. 
However,  the  effective  head,  or  any  of  the  individual  terms 
composing  it,  may  be  called  "head"  without  any  qualifying 
adjective. 

It  is  often  convenient  to  let  a  single  letter  stand  for  the  effective 
head,  hence  we  may  write 

H  =  p  +  z  +  ~-  (22) 

Using  this  brief  notation  we  may  rewrite  equation  (21)  as 

H!  -  H'  =  Hv  (23) 

We  see  that  the  effective  head  must  decrease  in  the  direction  of 
flow  by  an  amount  H'.  Hence,  although  either  pressure,  eleva- 


HYDROKINETICS 


53 


tion,  or  velocity  may  increase  in  the  direction  of  flow,  the  sum 
of  all  three  of  them  must  continually  decrease.  Therefore  an 
increase  in  one  of  these  items  must  always  be  accompanied  by  a 
corresponding  decrease  in  one  or  both  of  the  other  heads. 


EXAMPLES 

1.  Assuming  a  body  of  water  at  rest  in  Fig.  57,  so  that  there  is  no  loss 
of  head,  what  are  the  values  of  the  pressure  head  at  A,  B,  C,  and  D?     What 

are  the  values  of  the  elevation  head? 
What  are  the  values  of  the  effective 
head  at  these  four  points? 

2.  In  Fig.  58  the  point  A  is  30  ft. 
higher  than  B.  Assuming  the  pipe  to 
be  of  uniform  diameter,  in  which 


FIG.  57. 


FIG.  58. 


direction  will  the  water  be  flowing  if  the  pressure  at  A  is  20  Ib.  per  sq. 
in.  and  that  at  B  is  40  Ib.  per  sq.  in.?  What  is  the  head  lost  between 
the  two  points?  What  would  the  pressure  be  at  A  if  the  flow  were  to  be 
in  the  opposite  direction,  the  rate  of  discharge  remaining  the  same? 


FIG.  59. 


3.  In  Fig.  59  suppose  8  cu.  ft.  of  water  per  sec.  to  be  flowing  from  A  to 
C.    Assume  a  loss  of  head  from  A  to  B  to  be  equal  to  0.001  V*B  and  a  nequal 
loss  to  occur  between  B  and  C.     If  the  pressure  head  at  B  is  2  ft.,  how  high 
will  the  water  stand  in  piezometer  tubes  at  A  and  C? 

4.  Neglecting  all  loss  of  head  in  Fig.  55,  what  kind  of  a  curve  would  ex- 

the  variation  of  pressure  from  A  to  C? 


54  HYDRAULICS 

46.  Energy  and  Power  Meaning  of  Head. — Suppose  we 
multiply  the  elevation  head  z  by  the  weight  G  of  a  definite  volume 
of  water.  The  product  Gz  being  pounds  times  feet  represents 
foot-pounds  and  we  recognize  it  as  potential  energy.  That  is 
the  body  of  weight  G  possesses  Gz  foot-pounds  of  potential  energy 
by  reason  of  its  elevation  z.1  In  like  manner  if  we  multiply  the 
velocity  head  by  G  we  have  GV2/2g,  which  represents  the  kinetic 
energy  of  G  pounds  of  water  due  to  its  mass  and  velocity.  By 
analogy  we  might  expect  that  if  we  multiplied  the  pressure  head 
p  by  G  we  should  also  have  Gp  foot-pounds  of  energy  which  we 
could  call  pressure  energy. 

But  we  here  face  the  difficulty  that  we  recognize  energy  in 
only  two  fundamental  forms  which  we  call  potential  energy  or 
kinetic  energy  as  the  case  might  be.  All  other  forms  of  energy 
may  be  reduced  to  one  of  these  two.  It  is  not  clear  that  pressure 
energy  can  be  reduced  to  either  of  these  and  so  we  have  to  seek 
further  for  an  explanation  of  this  term.  Energy  is  ability  to  do 
work  and  we  feel  that  water  under  pressure  is  capable  of  doing 
work.  But  if  a  particle  of  water  should  in  some  manner  suddenly 
be  disconnected  from  its  fellows  it  would  still  have  its  initial 
elevation  and  velocity;  these  are  qualities  that  it  possesses  in  itself . 
But  its  pressure  would  be  lost,  since  that  is  derived  from  contact 
with  other  particles.  Thus  water  can  do  work  due  to  pressure  only 
so  long  as  it  is  still  connected  with  other  particles.  Hence  we 
might  conclude  that  pressure  energy,  if  the  term  is  permissible, 
is  not  something  that  a  particle  possesses  but  is  merely  energy 
that  is  transmitted  from  one  to  the  other  by  virtue  of  the  pressure 
and  the  motion. 

A  good  analogy  to  the  way  that  energy  may  be  transmitted 
past  any  point  is  offered  by  a  belt  connecting  two  pulleys.  The 
belt  possesses  kinetic  energy  due  to  its  mass  and  velocity  and 
this  energy  is  carried  past  any  stationary  reference  point.  But 
in  addition  to  this  the  belt  is  under  tension  and  is  in  motion  and 
hence  transmits  energy  from  one  pulley  to  the  other  that  it  does 
not  in  itself  possess.  In  like  manner  in  the  case  of  a  flowing 
stream  of  water,  the  water  carries  across  any  transverse  section 
a  certain  amount  of  energy  which  it  possesses  in  either  the 
potential  or  the  kinetic  form  or  both.  In  addition  to  this  it 

1  Strictly  speaking  this  energy  is  possessed  by  the  system  consisting  of 
the  earth  and  the  water. 


HYDROKINETICS  55 

may  transmit  energy  across  the  section  due  to  its  pressure  and 
motion. 

Suppose  we  consider  a  particle  of  water  flowing  from  (1)  to  (2) 
in  Fig.  56  and  assume  that  there  is  no  loss  of  head  so  that  H  = 
p  +  z  +  V2/2g  =  constant.  In  the  case  of  a  freely  falling  body 
acted  upon  by  no  other  forces  save  gravity  we  should  find  that 
the  loss  of  potential  energy  was  compensated  for  by  an  equal 
increase  in  kinetic  energy  so  that  its  energy  remains  constant. 
But  if  we  assume  that  the  stream  of  water  is  confined  in  a  channel 
of  uniform  area  the  velocity  is  constant  according  to  the  equation 
of  continuity,  hence  the  kinetic  energy  cannot  change.  If  the 
particle  loses  potential  energy  without  any  increase  in  its  kinetic 
energy,  it  follows  that  its  total  energy  must  decrease.  This  is 
true  for  the  particle.  But  the  total  energy  in  the  system  does 
not  change  for  we  are  assuming  no  loss.  The  reason  the  velocity 
of  the  particle  of  water  cannot  increase  is  that  negative  work  is 
being  done  upon  it  by  the  pressure  dp'dF  (Fig.  56).  But  this 
negative  work,  although  it  reduces  .the  energy  of  the  particle, 
is  not  lost  from  the  system.  If  conditions  permit  the  particle 
of  water  to  again  ascend  to  a  higher  elevation  or  to  increase  in 
velocity  the  pressure  acting  on  it  will  then  do  positive  work  and 
restore  the  potential  or  kinetic  energy  to  it. 

The  discussion  may  be  concluded  by  stating  that  head  repre- 
sents energy  per  unit  weight  of  water.  In  the  case  of  lost  head 
H'  represents  the  energy  lost  and  dissipated  in  the  form  of  heat 
per  unit  weight  of  water.  Thus  head  may  represent  foot- 
pounds per  pound  of  water. 

If  we  multiply  the  effective  head  H  by  the  weight  of  water  W 
flowing  across  any  section  per  unit  time,  the  product  of  the  two 
will  be  energy  per  unit  time.  But  energy  per  unit  time,  or  the 
rate  at  which  energy  is  transmitted,  is  power.  In  our  system 
of  units  the  product  WH  will  be  in  foot-pounds  per  second. 
Thus  head  is  equal  to  the  energy  per  unit  weight  of  water  or  it 
is  equal  to  the  power  per  unit  rate  of  discharge. 

EXAMPLES 

1.  The  surface  of  a  lake  is  500  ft.  above  a  certain  arbitrary  datum  plane 
with  respect  to  which  energy  is  to  be  measured,     (a)  What  is  the  energy 
per  pound  of  water?     (6)  If  the  lake  is  capable  of  furnishing  200  cu.  ft.  of 
water  per  sec.  what  power  is  available? 

2.  In  a  pipe  line  which  is  24  in.  in  diameter  we  have  water  flowing  with 
a  velocity  of  15  ft.  per  sec.  under  a  pressure  of  10  Ib.  per  sq.  in.     What  power 


56  HYDRAULICS 

is  being  transmitted  through  the  pipe  due  to  pressure?  What  is  the  total 
power  delivered? 

Ans.  123.2  hp.;  141.8  hp. 

3.  A  jet  of  water  free  from  all  pressure  is  7  in.  in  diameter  and  has  a  veloc- 
ity of  250  ft.  per  sec.     What  is  the  horsepower? 

Ans.  7,390  hp. 

4.  A  pipe  line  draws  water  from  a  lake  and  delivers  it  to  a  power  house 
at  a  point  500  ft.  below  the  level  of  the  surface  of  the  lake.     The  water  is 
delivered  at  a  velocity  of  170  ft.  per  sec.  by  a  jet  6  in.  in  diameter,  and  free 
from"  all  pressure  (save  that  of  the  atmosphere).     What  horsepower  has  been 
lost  in  the  pipe  line? 

47.  Correct  and  Incorrect  Applications  of  Bernoulli's  Theorem. 

— Bernoulli's  theorem  states  that  along  any  stream  line  the 
effective  head  remains  constant.  But  in  a  real  fluid  which  is 
viscous  there  can  be  no  flow  whatever  without  some  loss  due 
to  friction.  Hence  the  correct  statement  is  that  along  any 
stream  line  the  effective  head  always  decreases  in  the  direction 
of  flow. 

It  should  be  emphasized .  that  the  general  equation  should  be 
written  only  between  two  points  in  the  same  stream  line  so  that 
a  particle  of  water  may  be  assumed  to  flow  from  one  point  to 
the  other.  If  there  were  no  loss  .of  energy  it  would  follow  that  the 
effective  head  is  constant  at  all  points  throughout  a  connected 
body  of  fluid  and  in  that  event  only  we  might  apply  the  equation 
to  any  two  points  whatever.  But  in  reality  this  would  lead  us 
to  incorrect  conclusions  as  the  following  will  show. 

Suppose  that  we  have  no  loss  due  to  friction;  it  would  then 
follow  that  we  should  have  all  particles  of  water  at  a  section 
moving  with  equal  velocities  in  parallel  paths  as  shown  in  Fig.  50. 
All  particles  of  water  would  have  the  same  amount  of  kinetic 
energy  and  it  is  clear  that  all  particles  of  water  through  the  section 
would  have  the  same  amount  of  energy.  But  in  reality  the 
velocity  across  any  section  of  a  circular  pipe,  for  example,  is 
like  that  in  Fig.  51.  This  is  due  to  the  fact  that  the  greater 
frictional  resistance  near  the  walls  of  the  pipe  has  retarded  the 
water  near  them.  Certain  persons  have  incorrectly  applied 
Bernoulli's  theorem  between  a  point  near  the  wall  and  a  point  at 
the  center  of  the  pipe  and  reasoning  that  there  could  be  no  loss 
between  two  such  points  come  to  the  conclusion  that  the  pressure 
is  less  at  the  center  of  the  pipe  than  it  is  near  the  wall  because 
the  velocity  head  is  higher.  This  would  lead  to  an  excessive 
pressure  difference  if  true.  This  reasoning  has  even  been  bol- 


HYDROKINETICS  57 

stered  up  by  claims  of  experimental  evidence,  but  in  reality  the 
data  was  inaccurately  determined. 

It  is  no  more  permissible  to  apply  the  general  equation  be- 
tween two  points  in  adjacent  stream  lines  than  between  two 
separate  streams  in  different  channels.  It  is  a  mistake  to  assume 
that  the  effective  head  is  constant  across  any  section.  Correct 
experimental  evidence  shows  that  the  pressure  head  across  any 
section  varies  only  according  to  the  depth,  the  same  as  in  the 
case  of  water  at  rest.  Hence  if  the"  sum  of  pressure  head  and 
elevation  head  is  constant  across  any  section  while  the  velocity 
head  varies,  it  follows  that  the  total  head  varies  at  different 
points  in  the  section.  This  is  in  harmony  with  a  correct  appli- 
cation of  Bernoulli's  theorem  along  the  different  assumed  stream 
lines.  If  all  particles  started  with  the  same  store  of  energy  it 
is  clear  that  those  near  the  pipe  wall  would  have  lost  more  by 
friction  than  those  near  the  center.  Hence  the  energy  of  those 
particles  near  the  pipe  wall  should  be  less  than  that  of  those  in 
the  center. 

In  practical  application  of  the  general  equation  we  do  not 
deal  with  stream  lines  but  with  entire  streams.  Hence  we  have 
for  H  the  average  head  across  the  section  and  for  V  the  average 
velocity  across  the  section.  But  we  do  equate  the  average  head 
at  some  section  of  a  stream  to  the  average  head  at  some  other 
section  of  the  same  stream. 

In  considering  an  entire  stream,  rather  than  a  single  stream 
line,  we  assume  the  kinetic  energy  per  unit  time  to  be  WV2/2g, 
where  V  is  the  average  velocity.  This  is  not  strictly  true  for, 
if  the  velocity  varies  from  point  to  point  over  the  section,  the 
kinetic  energy  is  the  sum  of  the  kinetic  energies  of  all  the  in- 
dividual particles.  Considering  an  elementary  area  dF,  the  flow 
through  it  will  be  wV'dF,  where  V  is  the  actual  velocity  at  the 
point  in  question.  The  kinetic  energy  of  the  elementary  stream 
would  be  wV'3dF/2g.  Hence  the  total  kinetic  energy  for  the 
entire  stream  is 


-  Cv 
igj 


*dF.  (24) 

If  the  velocity  is  constant  this  will  become  wV3F/2g  =  WV2/2g 
since  the  true  velocity  at  every  point  is  the  average  velocity 
for  the  section.  But  in  reality  the  velocity  does  vary  to  some 
extent  over  the  section  and  hence  (24)  gives  the  true  kinetic 


58  HYDRAULICS 

energy.  If  the  law  of  variation  of  V  throughout  the  section  is 
known  this  integral  can  be  evaluated,  but  in  any  event  it  can  be 
shown  that  the  kinetic  energy  so  obtained  is  greater  than  that 
computed  by  using  the  average  velocity.1  Thus  making  the 
assumption  that  the  velocity  in  the  center  of  a  circular  pipe  is 
twice  that  near  the  walls  and  that  the  velocity  curve  is  a  semi- 
ellipse  it  will  be  found  that  the  true  kinetic  energy  is  1.06  times 
that  based  upon  the  mean  velocity.  Fortunately  the  difference 
is  not  great  in  important  cases  met  with  in  practice.  Thus  in 
the  case  of  a  jet  of  water  from  a  good  nozzle,  where  there  is 
little  variation  in  velocity,  the  difference  may  be  a  matter  of 
about  1  per  cent,  only.2  A  correct  application  of  equation 
(21)  would  require  us  to  insert  some  factor  before  the  velocity 
head,  based  upon  the  average  velocity,  to  give  a  correct  value. 
But  if  the  velocity  curves  at  sections  (1)  and  (2)  are  similar  and 
the  velocities  nearly  the  same  in  value  the  error  in  one  may 
nearly  balance  that  in  the  other.  Hence  it  is  not  customary  to 
allow  for  this  discrepancy  between  the  true  kinetic  energy  and 
that  computed  by  using  the  average  velocity. 

EXAMPLE 

1.  Assume  that  in  a  rectangular  stream  the  velocity  of  the  water  is  uni- 
form from  side  to  side  at  any  depth  but  that  it  varies  from  the  top  to  the 
bottom  inversely  as  the  depth.  If  the  velocity  at  the  top  is  twice  that  at 
the  bottom  find  the  ratio  between  the  true  kinetic  energy  passing  a  section 
per  unit  time  and  that  based  upon  the  mean  velocity. 

Ans.  1.11. 

48.  Applications  of  General  Equation. — For  the  solution  of 
problems  in  hydrokinetics  we  have  two  fundamental  equations, 
the  equation  of  continuity  (20)  and  the  general  equation  for 
steady  flow  (21),  usually  known  as  Bernoulli's  theorem.  In  most 
cases  thfc  following  procedure  may  be  employed: 

1.  Choose  a  datum  plane  through  any  convenient  point. 

2.  Note  at  what  sections  the  velocity  is  known  or  assumed. 
If  at  any  point  the  cross-section  is  great  as  compared  with  its 
value  elsewhere,  the  velocity  will  be  so  small  that  the  velocity 
head  may  be  disregarded. 

3.  Note  at  what  points  the  pressure  is  known  or   assumed. 
In  a  body  of  water  at  rest  with  a  free  surface  the  pressure  is 

1  L.  M.  Hoskins,  "Hydraulics,"  page  119. 

2  W.  R.  Eckart,  Jr.,  Inst.  of  Mech.  Eng.,  Jan.  7,  1910. 


HYDROKINETICS  59 

known  at  every  point.     The  pressure  in  a  jet  is  the  same  as  that 
in  the  medium  surrounding  the  jet. 

4.  Note  if  there  is  any  point  where  the  three  items  of  pressure, 
elevation,  and  velocity  are  known.  , 

5.  Note  if  there  is  any  point  where  there  is  only  one  unknown 
quantity. 

It  is  generally  possible  to  write  equation  (21)  between  two 
points  such  that  they  fulfill  conditions  (4)  and  (5)  respectively. 
Then  the  equation  may  be  solved  for  the  one  unknown.  If  it  is 
necessary  to  have  two  unknowns  then  equation  (21)  must  be 
solved  simultaneously  with  equation  (20).  The  procedure  is 
best  shown  by  applications  such  as  the  following: 


In  Fig.  60  we  have  a  pipe  BCD  which  is  6  in.  in  diameter 
through  which  water  flows  from  reservoir  A.  The  diameter  of 
the  stream  discharging  freely  into  the  air  at  E  is  3  in.  Let  us 
assume  that  n  =  2  in  equation  (21)  so  that  the  loss  of  head  due 
to  friction  is  proportional  to  the  square  of  the  velocity.  Then 
Hf  =  kV2/2g,  where  V  is  the  velocity  in  the  pipe.  Suppose  that 
the  roughness  of  the  pipe  and  the  lengths  between  the  various 
points  are  such  that  the  values  of  k  from  the  reservoir  to  B, 
from  B  to  C,  from  C  to  D,  and  from  D  to  E  are  2,  4,  4,  and  1 
respectively.  Let  it  be  required  to  find  the  pressure  at  C  when 
flow  takes  place. 

At  C  there  is  both  an  unknown  pressure  and  an  unknown 
velocity,  hence  we  cannot  immediately  apply  ^  equation  (21)  as 
one  equation  is  capable  of  determining  only  one  unknown.  Let 
us  then  follow  the  procedure  outlined.  The  location  of  a  datum 
plane  is  immaterial  in  the  solution  of  the  problem  but  it  is  usually 
convenient  to  take  it  through  the  lowest  point  in  the  figure 
and  thus  avoid  negative  values  of  z.  Therefore  let  us  assume 
a  datum  plane  through  E.  In  the  reservoir  we  find  that  the 
velocity  is  negligible  because  of  the  large  area  as  compared  with 


60  HYDRAULICS 

the  area  of  the  pipe.  At  a  point  A  on  the  surface  of  the  water 
we  find  the  pressure  to  be  atmospheric,  which  is  also  the  case 
with  the  stream  at  E.  Thus  whatever  the  pressure  of  the  atmos- 
phere may  be  its  effects  can  easily  be  shown  to  balance  out  and 
therefore  we  neglect  it  altogether.  Hence  at  A  we  find  that 
everything  is  known  while  at  E  the  velocity  head  is  the  only 
unknown. 

We  shall  apply  equation  (21),  or  its  equivalent  (23),  between 
points  A  and  E.  We  find  that 

HA        =0  +  40  +  0 
HE        =  0  +  0  +  F'2/2<7 
#'A_E  =  11  F2/20. 

Now  V  is  the  velocity  of  the  jet  at  E  while  V  is  the  velocity 
in  the  pipe  but  one  may  be  replaced  in  term  of  the  other  by 
equation  (20).  (It  is  seldom  necessary  to  compute  areas  for 
this.  It  is  both  easier  and  more  accurate  to  use  the  ratios  of 
the  areas,  which  means  the  ratios  of  the  diameters  squared.) 
Now  V  =  FV/F',  where  F'  is  the  area  of  the  jet.  But  F/F'  = 
(6/3)2  =  22  =  4.  Hence  V  =  4V  and  V'2  =  16F2.  Replacing 
V  and  V  and  substituting  in  equation  (23)  we  have 

40  -  llF2/20  =  16F2/2#. 

Thus  F2/20  =  40/27  =  1.48  ft.  We  have  now  determined  one 
of  the  unknowns  at  C. 

We  may  next  apply  equation  (23)  between  C  and  either  A  or 
E  since  we  know  the  value  of  H  at  either  of  the  latter  points. 
The  value  of  the  effective  head  at  C  is  H  =  p  +  25  +  1.48 
while  tf'A_c  =  6F2/20  =  6  X  1.48  =  8.88  ft.  Now  from  (23) 

40  -  8.88  =  Hc  =  P  +  26.48. 
Hence  p  =  4.64  ft. 

If  the  rate  of  discharge  is  also  desired  we  can  easily  find  that. 
Since  F2/20  =  1.48,  F  =  V2g  1.48  =  8.025\/L48  =  9.78  ft. 
per  sec.  Hence  q  =  0.196  X  9.78  =  1.92  cu.  ft.  per  sec. 

EXAMPLES 

1.  Compute  the  pressures  at  B  and  D  in  Fig.  60. 

2.  Suppose  that  all  other  data  for  Fig.  60  remains  unchanged  except  the 
diameter  at  C.     What  will  this  diameter  be  if  there  is  a  vacuum  of  20  in. 
of  mercury  at  C? 


HYDROKINETICS 


61 


3.  Suppose  the  diameter  at  C  in  Fig.  60  remains  6  in.  and  all  other  data 
is  likewise  unchanged  except  the  elevation  of  C.     How  far  above  E  can  C 
be  placed  to  produce  a  vacuum  of  20  in.  of  mercury? 

4.  What  fall  from  the  surface  of  the  reservoir  in  Fig.  60  to  the  outlet  at 
E  would  be  necessary  to  produce  the  same  rate  of  discharge,  if  there  were 
no  loss  due  to  friction? 

49.  PROBLEMS 

1.  In  Fig.  60,  with  data  as  given  in  preceding  article,  what  is  the  energy 
per  cubic  foot  of  water  in  the  reservoir?     What  is  the  power  transmitted 
past  C?     What  is  the  power  in  the  jet  at  E?     What  is  the  value  of  the  power 
lost  by  friction? 

2.  Suppose  all  radiation  of  heat  from  the  pipe  in  Fig.  60  could  be  pre- 
vented and  that  the  temperature  of  the  water  in  the  reservoir  is  32°F. 
What  would  be  the  temperature  of  the  water  in  the  jet?     (778  ft.-lb.  of  work 
will  raise  1  Ib.  of  water  1°F.) 


FIG.  61. 


3.  In  the  siphon  shown  in  Fig.  61  the  loss  of  head  from  the  intake  to  B 
is  4  ft.  and  that  from  B  to  the  discharge  end  of  the  pipe  is  3  ft.     Find  the 
rate  of  discharge  and  the  pressure  head  at  B  if  the  pipe  is  of  a  uniform  diame- 
ter of  6  in. 

Arts,  p  =  —  14  ft. 

4.  Suppose  the  discharge  end  of  the  siphon  of  Fig.  61  were  4  in.  in  diameter, 
other  data  remaining  the  same,  what  would  be  the  rate  of  discharge  and 
pressure  at  B? 

5.  If  all  data  were  as  given  in  problem  (4)  except  that  the  size  of  the 
stream  discharging  from  the  end  of  the  siphon  were  not  fixed,  how  large 
could  the  diameter  of  this  be  if  the  pressure  at  B  is  —25  ft.?     What  would 
then  be  the  rate  of  discharge? 

6.  The  diameter  of  the  pipe  in  Fig.  62  is  4  in.  and  that  of  the  stream  dis- 
charging into  the  air  at  E  is  3  in.     Neglecting  all  losses  of  energy,  what  are 
the  pressures  at  B,  C,  and  D?     (Velocity  assumed  negligible  at  A.) 


CHAPTER  VI 
APPLICATIONS  OF  HYDROKINETICS 

50.  Definition  of  a  Jet. — A  jet  is  a  stream  bounded  by  a  fluid 
of  a  different  kind.     The  jets  with  which  we  are  concerned  in 
practical  hydraulics  are  streams  of  water  entirely  surrounded  by 
air.     It  is  evident  that  the  pressure  to  which  the  water  in  a  jet 
is  subjected  is  exactly  equal  to  the  pressure  exerted  upon  its 
boundaries  by  the  surrounding  air. 

51.  Jet  Coefficients. — Cue  to  frictional  resistance  the  actual 
velocity  of  a  jet  is  always  less  than  would  otherwise  be  the  case. 


From  a  photograph  by  W.  R.  Eckart,  Jr. 

FIG.  63. —  Jet  from  7^-inch  nozzle.     (Head  =  822  ft.,  velocity  =  227.4  ft. 

per  sec.) 

The  velocity  which  would  be  attained  if  friction  did  not  exist 
may  be  termed  the  ideal  velocity.1  The  ratio  of  the  actual 
velocity  to  the  ideal  velocity  is  called  the  coefficient  of  velocity. 

1  This  is  frequently  called  "theoretical  velocity"  by  others  but  the  author 
feels  that  this  is  a  misuse  of  the  word  "theoretical."  Any  correct  and 
sensible  theory  should  allow  for  the  fact  that  friction  exists  and  affects  the 
result.  Otherwise  it  is  not  theory  but  merely  an  incorrect  hypothesis. 

62 


APPLICATIONS  OF  HYDROKINETICS 


63 


The  area  of  the  opening  through  which  the  jet  issues  is  some- 
thing that  is  readily  determined,  but  in  many  cases  the  area 
of  the  jet  cannot  so  readily  be  measured  without  special  equip- 
ment. Hence  it  is  desirable  to  know  the  relation  between  the 
area  of  a  jet  and  the  area  of  the  open- 
ing through  which  it  came.  This 
factor,  the  ratio  of  the  area  of  the 
jet  to  the  area  of  the  opening,  is  called 
the  coefficient  of  contraction.  The 
word  "contraction"  is  used  because 
the  jet  usually  contracts  and  is 
smaller  than  the  opening,  as  may  be 

seen  in  Fig.  63.     In  case  the  jet  does 

•    .  FIG.  64. — The  "vena 

contract,    the    section    of    minimum  contracta." 

area,   the   "vena   contracta,"   is  the 

section  whose  area  is  considered  in  the  calculations.    The  velocity 

of  a  jet  is  also  understood  to  be  the  velocity  found  at  this  point. 

The  coefficient  of  contraction  may  be  unity,  indicating  that 

the  area  of  the  jet  is  equal  to  the  area  of  the  opening  from  which 


From  a  photograph  by  the  author. 
FIG.  65.— Discharge  from  end  of  a  straight  pipe.     (Mixture  of  water,  mud, 
and  rocks  building  up  the  Calaveras  earth  dam.) 

it  issued.  ,This  is  the  case  when  the  sides  of  the  stream  are 
parallel  before  it  issues  from  the  opening,  as  when  the  discharge 
is  from  the  open  end  of  a  pipe  as  shown  in  Fig.  65.  Of  course 
after  passing  the  point  of  minimum  section  the  jet  diverges  again 


64 


HYDRAULICS 


due  to  the  loss  of  velocity  from  frictional  resistance, 
seen  in  Figs.  63,  66,  and  67. 


This  is 


From  a  photograph  by  the  author. 

FIG.  66. — Jet  from  hydraulic  giant  sluicing  out  material    for  Calaveras 

earth  dam. 


From  a  photograph  by  the  author. 

FIG.  67. — Nearer  view  of  jet  in  preceding  figure. 

The  product  of  the  coefficient  of  velocity  and  the  coefficient  of 
contraction  is  called  the  coefficient  of  discharge.  It  is  the  ratio 
of  the  actual  rate  of  discharge  to  the  ideal  rate  of  discharge 


APPLICATIONS  OF  HYDROKINETICS 


65 


that  would  be  obtained  if  there  were  no  friction  and  if  the  jet 
did  not  contract. 

52.  Flow  through  Orifice. — An  orifice  is  any  opening  in  the 
wall  of  a  containing  vessel.  The  only  restriction  is  that  the 
thickness  of  the  wall  shall  be  only  a  small  fraction  of  the  diam- 
eter or  other  linear  dimension  of  the  opening. 

Let  us  write  the  general  equation  (23)  between  points  (1)  and 
(2)  of  Fig.  68,  assuming  the  pressure  is  atmospheric  at  both 
points  and  that  the  area  of  the  vessel  is  such  that  the  velocity 


FIG.  68. 


at  (1)  may  be  neglected.  We  shall  also  assume  a  loss  of  head 
between  the  two  points  which  we  shall  consider  as  being  propor- 
tional to  the  square  of  the  velocity  of  the  jet  V  and  introduce 
such  a  factor  k  that  H  'i_2  =  kV2/2g.  Thus  we  have 


0 


0, 


V2/2g. 


Hence  Hl  -  kV2/2g  =  V2/2g. 
From  this,  V2/2g  =  #,/(!  +  fc)  or, 


(25) 


If  there  were  no  frictional  resistance  to  flow,  the  value  of  k 
would  be  zero.  Thus  the  ideal  velocity  is  V  =  -\/2gHi.  It  is 
the  ideal  velocity  that  is  to  be  multiplied  by  the  coefficient  of 
velocity  to  obtain  the  true  velocity.  Hence 

(26) 


66 


HYDRAULICS 


In  this  equation  we  have  the  relation  between  the  coefficient  of 
velocity  and  the  coefficient  of  loss.  By  squaring  both  sides  and 
rearranging  this  may  also  be  written 


k 


(27) 


Since  in  this  case  HI  =  h,  we  may  now  write  equation  (25) 
in  its  more  usual  and  more  convenient  form 


V  =  cv 


(28) 


53.  Orifice  in  Case  of  Unequal  Pressures.  —  In  case  the  jet 
in  Fig.  68  discharged  into  a  medium  under  a  different  pressure 
from  that  existing  upon  the  surface  of  the  liquid  in  the  vessel 
we  should  proceed  as  follows: 

ff  i  =  Pi  +  h  +  0,     #2  =  p2  +  0  +  V2/2g 

Completing  the  solution  as  in  the  preceding  article  we  should 
obtain 

V  =  cvV2g(h  +  pi-  pz)  (29) 

54.  Submerged  Orifice.  —  For  a  submerged  orifice  as  shown  in 
Fig.  69  we  could  write  for  points  (1)  and  (2)  : 

#1  =  0  +  (h  +  y)  +  0,     #2  =  y  +  0  +  F2/2<7. 


FIG.  69. 

The  pressure  in  the  stream  of  water  at  (2)  is  equal  to  that  upon 
its  boundaries  and  that  is  equal  to  the  depth  y.  It  is  evident 
that  y  cancels  out  so  that  for  the  submerged  orifice  we  also  have 

V  =  cv\/2gh  (30) 

The  coefficients  for  a  submerged  orifice  would  be  different 
from  those  for  an  orifice  discharging  into  the  air.  It  is  possible 
that  the  contraction  coefficient  would  be  materially  larger  and 
the  velocity  coefficient  somewhat  smaller. 


APPLICATIONS  OF  HYDROKINETICS 


67 


EXAMPLES 

1.  In  Fig.  69  determine  the  loss  of  head  between  (2)  and  (3),  the  velocity 
at  the  latter  point  being  negligible.    Ans.  V2/2g. 

2.  Knowing  the  loss  of  head  between  (2)  and  (3)  add  this  to  kVz/2g 
and  solve  for  the  velocity  of  flow  through  the  orifice  by  writing  an  equa- 
tion between  (1)  and  (3). 

55.  Values  of  Orifice  Coefficients. — The  value  of  the  coefficient 
of  velocity  is  always  less  than  unity,  though  it  may  often  approach 
that  value  very  closely.  In  some  very  well  made  nozzles  and 
sharp-edged  orifices  the  coefficient  of  velocity  may  be  as  high  as 
0.98  and  occasionally  0.99  may  be  attained. 


(6)  (<) 

FIG.  70. — Types  of  orifices. 

A  standard  orifice  is  one  with  a  sharp  edge  as  in  Fig.  70  (a). 
It  is  called  a  standard  orifice  because  one  will  give  practically 
the  same  results  as  another  of  the  same  size.  Any  other  form 
of  orifice  such  as  (c)  in  Fig.  70  would  give  different  results  de- 
pending upon  the  thickness  of  the  plate,  the  roughness  of  the 
material,  etc.;  hence  the  coefficients  for  each  individual  one 
would  have  to  be  determined  if  accurate  computations  were 
desired. 

If  the  plate  is  thin  and  the  inner  corner  is  square  and  sharp, 
the  orifice  in  Fig.  70  (b)  may  also  be  considered  a  standard 
orifice.  But  if  the  plate  is  too  thick  the  condition  in  Fig.  70  (c) 
is  met  with  and  the  velocity  coefficient  will  be  less  than  in  the 
former  case  due  to  the  greater  frictional  resistance  the  water 
encounters.  Rounding  the  edge  as  in  Fig.  70  (d)  reduces  the 
eddy  losses  and  hence  increases  the  coefficient  of  velocity  slightly. 

The  contraction  coefficient  is  much  more  sensitive  to  varia- 
tions in  the  nature  of  the  orifice  than  is  the  velocity  coefficient. 
Buf  it  should  be  noted  that  contraction  affects  only  the  size  of 


68  HYDRAULICS 

the  jet  and  not  its  velocity.  The  coefficient  of  contraction  is  the 
least  in  (a)  and  (b)  and  may  be  unity  in  (c)  and  (d).  Hence  the 
latter  forms  of  orifices  may  discharge  much  more  water  than  the 
former.  Thus  the  type  of  orifice  to  be  used  depends  upon 
whether  one  wishes  a  large  discharge  or  wishes  the  maximum 
velocity.  Of  course  if  an  orifice  is  used  as  a  means  of  measur- 
ing rate  of  discharge  only  the  standard  orifice  should  be  em- 
ployed unless  the  special  orifice  can  be  calibrated. 

If  F  and  V  denote  the  area  and  velocity  of  the  jet  respec- 
tively, while  F0  is  the  area  of  the  orifice,  it  may  be  seen  that, 
since  c  =  cccv, 

q  =  FV  =  (ccFoKcvV^gHl)  =  cF0V2gH[. 

Values  of  coefficients  of  discharge  for  standard  circular  and 
square  orifices  are  given  in  Tables  1  and  2  respectively.  These 
orifices  are  to  be  sharp  edged  and  to  be  in  vertical  planes.  The 
orifice -should  be  so  situated  that  it  is  flush  with  a  flat  wall  on  the 
water  side  free  from  all  obstructions,  projections,  and  sides 
for  a  distance  in  all  directions  of  at  least  three  times  the  diameter 
of  the  orifice.  If  this  is  not  so  the  full  contraction  will  not  be 
obtained  and  the  actual  coefficient  will  be  larger  than  given  by 
the  tables.  The  tables  show  that  the  coefficients  are  different  for 
different  sizes  and  for  a  given  orifice  vary  with  the  head  on  the 
orifice,  thus  illustrating  the  impossibility  of  stating  general 
values  or  laws  that  hold  in  all  cases.  It  will  be  noted  that  the 
coefficients  decrease  as  the  head  increases  and  that  as  higher 
values  are  attained  the  rate  of  decrease  is  much  smaller.  For 
heads  above  100  ft.  the  values  for  100  ft.  may  doubtless  be  ap- 
plied with  little  error. 

EXAMPLES 

1.  Find  values  of  the  coefficient  of  loss  when  the  velocity  coefficient  has 
values  of  1.00,  0.99,  0.95  and  0.80.     Find  cv  when  k  has  values  of  0,  0.5,  1.00. 

2.  Water  issues  from  a  vertical  orifice  (one  in  a  vertical  plane)  under  a 
head  of  16  ft.     The  diameter  of  the  orifice  =  2  in.     When  measured  it  was 
found  that  q  =  33  cu.  ft.  per  min.     What  is  the  coefficient  of  discharge?     If 
the  coefficient  of  velocity  is  assumed  to  be  0.96,  what  is  the  value  of  the 
coefficient  of  contraction?     What  will  be  the  diameter  of  the  jet? 

3.  The  discharge  from  an  orifice  under  a  head  of  230  ft.  was  found  to  be 
180  cu.  ft.  per  min.     The  jet  was  found  to  be  2.16  in.  in  diameter  while  the 
diameter  of  the  circular  orifice  was  2.25  in.     What  are  the  coefficients  of 
velocity,  contraction,  and  discharge? 

4.  What  would  be  the  rate  of  discharge  from  a  standard  circular  orifice 
of  2  in.  diameter  under  a  head  of  3  ft.?     What  would  be  the  probable 
value  of  the  jet  velocity? 


APPLICATIONS  OF  HYDROKINETICS 


69 


TABLE  I. — DISCHARGE  COEFFICIENTS  FOR  STANDARD  CIRCULAR  VERTICAL 
ORIFICES  ACCORDING  TO  HAMILTON  SMITH 


Head  on 
center  of 

Diamete 

r  of  orifie 

e  in  feet 

orifice  in 
feet 

0.02 

0.03 

0.04 

0.05 

0.07 

0.10 

0.20 

0.60 

1.00 

0.4 

0.637 

0.631 

0.624 

0.618 

0.6 

0.655 

0.640 

0.630 

0.624 

0.618 

0.613 

0.601 

0.593 

0.8 

0.648 

0.634 

0.626 

0.620 

0.615 

0.610 

0.601 

0.594 

0.590 

1.0 

0.644 

0.631 

0.623 

0.617 

0.612 

0.608 

0.600 

0.596 

0.591 

1.4 

0.638 

0.625 

0.618 

0.613 

0.609 

0.605 

0.600 

0.595 

0.693 

2.0 

0.632 

0.621 

0.614 

0.610 

0.607 

0.604 

0.599 

0.597 

0.595 

2.5 

0.629 

0.619 

0.612 

0.608 

0.605 

0.603 

0.599 

0.598 

0.596 

3.0 

0.627 

0.617 

0.611 

0.606 

0.604 

0.603 

0.599 

0.598 

0.597 

3.5 

0.625 

0.616 

0.610 

0.606 

0.604 

0.602 

0.599 

0.598 

0.596 

4.0 

0.623 

0.614 

0.609 

0.605 

0.603 

0.602 

0.599 

0.597 

0.596 

6.0 

0.618 

0.611 

0.607 

0.604 

0.602 

0.600 

0.598 

0.597 

0.596 

8.0 

0.614 

0.608 

0.605 

0.603 

0.601 

0.600 

0.598 

0.596 

0.596 

10.0 

0.611 

0.606 

0.603 

0.601 

0.599 

0.598 

0.597 

0.596 

0.595 

20.0 

0.601 

0.600 

0.599 

0.598 

0.597 

0.596 

0.596 

0.596 

0.594 

50.0 

0.596 

0.596 

0.595 

0.595 

0.594 

0.594 

0.594 

0.594 

0.593 

100.  0 

0.593 

0.593 

0.592 

0.592 

0.592 

0.592 

0.592 

0.592 

0.592 

TABLE  II. — DISCHARGE  COEFFICIENTS  FOR  STANDARD  SQUARE  VERTICAL 
ORIFICES  ACCORDING  TO  HAMILTON  SMITH 


Head  on  center  of 


Side  of  square  in  feet 


orifice  in  feet 

0.02 

0.04 

0.07 

0.10 

0.20 

0.60 

1.00 

0.4 

0.643 

0.628 

0.621 

0.6 

0.660 

0.636 

0.623 

0.617 

0.605 

0.598 

0.8 

0.652 

0.631 

0.620 

0.615 

0.605 

0.600 

0.597 

1.0 

0.648 

0.628 

0.618 

0.613 

0.605 

0.601 

0.699 

1.4 

0.642 

0.623 

0.614 

0.610 

0.605 

0.601 

0.698 

2.0 

0.637 

0.619 

0.613 

0.608 

0.605 

0.604 

0.602 

2.5 

0.634 

0.617 

0.610 

0.607 

0.605 

0.604 

0.602 

3.0 

0.632 

0.616 

0.609 

0.607 

0.605 

0.604 

0.603 

3.5 

0.630 

0.615 

0.609 

0.607 

0.605 

0.604 

0.602 

4.0 

0.628 

0.614 

0.608 

0.606 

0.605 

0.603 

0.602 

6.0 

0.623 

0.612 

0.607 

0.605 

0.604 

0.603 

0.602 

8.0 

0.619 

0.610 

0.606 

0.605 

0.604 

0.603 

0.602 

10.0 

0.616 

0.608 

0.605 

0.604 

0.603 

0.602 

0.601 

20.0 

0.606 

0.604 

0.602 

0.602 

0.602 

0.601 

0.600 

50.0 

0.602 

0.601 

0.601 

0.600 

0.600 

0.600 

0.599 

100.0 

0.599 

0.598 

0.598 

0.598 

0.598 

0.598 

0.598 

70 


HYDRAULICS 


56.  Flow  through  Short  Tubes. — A  short  tube,  whose  length 
is  not  more  than  two  or  three  diameters,  may  be  treated  in  the 
same  manner  as  the  orifice.  If  its  length  is  much  greater  it  be- 
comes a  pipe,  the  consideration  of  which  will  be  taken  up  in  a  sub- 
sequent chapter.  As  a  measuring  device  the  tube  is  not  as 
valuable  as  the  standard  orifice,  since  the  coefficients  of  the  latter 
are  known  with  greater  accuracy.  It  is  possible  to  apply  to 
any  standard  orifice  coefficients  taken  from  tables  and  to  have 
some  assurance  as  to  the  accuracy  of  the  result.  But  the  tube 


(<*)  (6) 

FIG.  71. — Coefficients  for  tubes. 

cannot  be  standardized  as  accurately  as  the  orifice  and  hence  it  is 
necessary  to  calibrate  the  tube  itself  if  it  is*  to  be  used  for  water 
measuring. 

The  coefficient  of  velocity  is  the  highest  in  the  case  of  the  bell- 
mouthed  tube,  Fig.  71  (a),  since  the  eddy  losses  at  entrance  are 
reduced  to  a  minimum.  The  greatest  hydraulic  friction  loss  is 
met  with  in  the  case  of  the  re-entrant  tube  in  Fig.  71(c) ;  hence  its 
velocity  coefficient  is  the  least.  The  contraction  coefficients  of 
all  tubes  are  unity,  provided  the  tubes  have  parallel  sides. 

EXAMPLES 

1.  In  Fig.  72  the  diameter  of  the  jet  at  Cis 
1.5  in.      If  the  coefficient  of  discharge  is  0.96, 
what  is  the  rate  of  discharge? 

2.  At  B  in  Fig.  72  the  diameter  is  1.0  in.  while 
that  at  C  is  1.5  in.     What  is  the  value  of  the 
velocity  and  pressure  at  Bt     (Assume  no  loss 
of  head  between  B  and  C.) 

3.  In  Fig.  72  if  the  pressure  at  B  is  to  be 
absolute  zero,  what  will  be  the  velocity  at  that 
point?     What  would  be  the  velocity  at  B  if 


FIG.  72. 


the  tube  were  cut  off  at  that  section? 


57.  Flow  through   Nozzles. — A  nozzle  is  a  converging   tube 
usually  placed  on  the  end  of  a  pipe  line  or  hose.     It  may  be  a 


APPLICATIONS  OF  HYDROKINETICS 


71 


plain  conical  nozzle  as  in  Fig.  73  (a)  or  a  smooth  convex  nozzle 
as  shown  in  Fig.  73 (c).     The  jet  from  a  nozzle  may  undergo  some 


(a)  Conical  Nozzle  (6)  Straight  Tip  (C)  Convex  Nozzle 

FIG.  73. — Standard  Nozzles. 


From  a  photograph  by  the  author. 
FIG.  74. — Jet  from  hydraulic  giant  washing  out  material  for  earth  fill  dam. 

contraction  or,  if  a  small  portion  near  the  mouth  is  of  uniform 
diameter  as  in  Fig.  73(6),  the  water  may  leave  in  parallel  lines 
and  suffer  no  contraction. 


72 


HYDRAULICS 


Nozzles  may  be  used  as  water-measuring  devices  the  same  as 
standard  orifices,  and  are  especially  useful  for  that  purpose  when 
high  heads  are  employed.  They  may  also  be  used  to  furnish  jets 
at  high  velocities  for  fire  purposes,  for  power,  or  for  hydraulic 
mining  and  similar  work  such  as  shown  in  Fig.  74. 

In  the  case  of  a  nozzle  the  water  in  the  pipe  leading  to  it 
usually  flows  with  a  velocity  which  is  quite  appreciable  and  it  is 
desirable  to  consider  this  velocity  of  approach.  Hence  we  shall 
derive  an  expression  which  takes  this  velocity  head  into  account. 


FIG.  75. 

For  points  (1)  and  (2)  of  Fig.  75  we  have  Hl  =  pl  +  0  -f  Vi*/2g, 
Hz  =  0  +  0  +  F22/2<7,  and  HV2=  kV^/2g.  Now  applying  the 
general  equation  (23)  we  have 

HI  —  kV^/2g  =  Vz2/2g. 
From  this 

F2  =  ^/Y+k  VtyHi  =  cv\/2gHl  (31) 

This  is  similar  to  the  result  for  the  orifice,  but  here  we  have 
HI  =  pi  +  Vi2/2g  instead  of  =  h  as  in  Art.  52.  Inserting  this 
value  we  have 

J  ~.  T7-     <>* 

(32) 

If  the  velocity  in  the  pipe  is  known  in  some  way  this  equation 
could  be  used  directly.  In  case  it  is  not  known,  we  may  proceed 
as  follows : 

Squaring  both  sides  of  equation  (32)  we  have 


.     Substituting 


By  the  equation  of  continuity   V\  = 
this  in  the  above  and  transposing  we  have 


APPLICATIONS  OF  HYDROKINETICS  73 

From  this  we  obtain 

F2  = 


In  this  equation  F2  is  the  actual  area  of  the  jet  at  the  point  of 
minimum  area.  If  the  area  of  the  nozzle  opening  be  expressed 
by  Fo,  then  F2  =  ccF0  and  the  latter  expression  could  be  inserted  in 
(33).  The  rate  of  discharge  is  obtained  by  multiplying  F2  by  Ft 
or  by  ccF0  as  the  case  may  be. 

The  velocity  coefficients  of  well-made  nozzles  are  very  high, 
being  practically  equal  to  those  of  a  standard  circular  orifice. 
We  may  reasonably  assume  an  average  value  of  the  velocity 
coefficient  of  0.98,  though  even  this  is  often  exceeded.1 

The  height  to  which  a  good  fire  stream  can  be  thrown  by  a 
nozzle  is  from  about  two-thirds  to  three-fourths  of  the  effective 
head  at  the  base  of  the  nozzle.  The  proportion  is  higher  for  large 
jets  than  for  small  ones,  for  smooth  nozzles  than  rough  ones,  and 
for  low  pressures  than  for  high  pressures. 

58.  Efficiency  of  Nozzle. — Since  a  nozzle  is  frequently  em- 
ployed for  power  purposes,  we  may  be  interested  in  its  efficiency. 
The  efficiency  of  a  nozzle  may  be  defined  as  the  ratio  of  the  power 
in  the  jet  to  the  power  delivered  to  the  nozzle.  But  we  have  seen 
that  for  a  given  rate  of  discharge,  power  is  directly  proportional 
to  head.  Thus  referring  to  Fig.  75,  we  have 

e  =  #2/#i. 

But  from  equation  (31)  H2  =  V2z/2g  =  cvzH^.  From  this  it  may 
be  seen  that 

e  =  c,2  (34) 

This  would  be  exactly  true  if  all  particles  of  water  in  the  jet 
possessed  the  same  velocity  and  hence  the  same  kinetic  energy. 
Actually  F2  is  the  average  velocity  of  the  jet,  and  it  has 
been  stated  in  Art.  47  that  the  true  kinetic  energy  of  a  stream  is 
greater  than  that  obtained  by  using  the  square  of  the  average 
velocity.  Hence  the  true  efficiency  of  a  jet  from  a  good  nozzle 
may  be  about  1  per  cent,  more  than  the  value  given  by  (34). 

1  John  R.  Freeman,  Trans.  A.  S.  C.  E.,  vol.  21,  page  303  (1889);  Trans. 
A.  S.  C.  E.,  vol.  24,  page  492  (1891). 

W.  R.  Eckart,  Jr.,  Inst.  of  Mech.  Eng.,  Jan.  7,  1910. 

V.  R.  Fleming,  Proc.  of  Fifth  Meeting  of  111.  Water  Supply  Assoc.,  1913. 

R.  L.  Daugherty,  "Hydraulic  Turbines,"  page  69. 


74 


HYDRAULICS 


EXAMPLES 

1.  In  Fig.  63  the  actual  measured  diameter  of  the  minimum  section  of 
the  j  et  was  6  *  %  Q  in. ,  the  area  of  the  nozzle  opening  being  43. 02  sq .  in .    Co  m  - 
pute  the  coefficients  of  velocity,  contraction,  and  discharge  using  the  values 
of  HI  and  F2  given.     What  is  the  efficiency  of  the  nozzle?     What  is  the 
horsepower  in  the  jet? 

2.  What  is  the  value  of  the  head  lost  in  hydraulic  friction  in  the  nozzle  of 
Fig.  63?     What  is  the  value  of  fc? 

3.  The  velocity  of  water  in  a  6-in.  pipe  is  12  ft.  per  sec.     At  the  end  of  the 
pipe  is  a  nozzle  whose  velocity  coefficient  is  0.98.     If  the  pressure  in  the  pipe 
at  the  base  of  the  nozzle  is  10  Ib.  per  sq.  in.,  what  is  the  velocity  of  the  jet? 
What  is  the  diameter  of  the  jet?     What  is  the  rate  of  discharge? 

4.  A  jet  2  in.  in  diameter  is  discharged  through  a  nozzle  whose  velocity 
coefficient  is  0.98.     In  the  pipe  at  the  base  of  the  nozzle  there  is  a  pressure 
of  10  Ib.  per  sq.  in.,  the  diameter  of  the  pipe  at  that  point  being  6  in.     What  is 
the  velocity  of  the  jet?     What  is  the  rate  of  discharge? 

6.  If  the  diameter  of  the  jet  in  problem  (4)  were  1.0  in.,  all  other  data 
remaining  the  same,  find  the  jet  velocity  and  the  rate  of  discharge. 

59.  Venturi  Meter. — If  water  is  caused  to  flow  through  the 
device  shown  in  Fig.  76,  the  increased  velocity  through  the 
"throat"  will  produce  a  corresponding  pressure  drop.  This  drop 
in  pressure  may  be  made  to  serve  as  a  measure  of  the  rate  of 
discharge.  Such  an  instrument  is  called  a  Venturi  meter. 


c 

D  • 

A. 

FIG.  76. — Venturi  meter. 

It  may  be  seen  that  the  Venturi  meter  is  very  similar  in  prin- 
ciple to  the  nozzle.  In  both  there  is  an  increase  in  velocity  of  the 
water  accompanied  by  a  corresponding  pressure  drop.  And  in 
both  the  rate  of  discharge  may  be  found  to  be  a  function  of  the 
pressure  drop.  The  only  difference  is  that  the  pressure  at  the 
throat  of  the  Venturi  meter  may  be  either  somewhat  greater  or 
less  than  atmospheric,  and  the  stream  at  that  point  is  not  a  free  jet 
but  is  expanded  again  to  fill  the  pipe  below  the  meter.  Hence  the 


APPLICATIONS  OF  HYDROKINETICS  75 

equations  for  the  nozzle  would  seem  to  apply  directly  to  the 
Venturi  meter,  with  pi  equal  to  the  pressure  drop  in  both  cases. 

Actually  the  coefficients  for  the  Venturi  meter  are  based  upon  a 
formula  derived  by  a  slightly  different  procedure.  Thus  we 
equate  HI  to  H  2  assuming  that  H'  is  zero,  and  then  introduce  a 
coefficient  as  the  very  last  step.  Assuming  the  meter  to  be  hori- 
zontal so  that  Zi  =  32,  we  have 

Pl  +  VS/2g  =  pz  +  F22/2<7. 
From  this, 

TV      TV 


the  equation  of  continuity  Vi  =  (F2/Fi)  72,  and  hence 


2gh 

V  2  = 


Since  there  is  some  slight  loss  of  head  between  (1)  and  (2)  the 
true  velocity  will  be  less  than  this  and  so  we  multiply  it  by  a 
velocity  coefficient.  We  then  have 


(35) 


This  may  be  seen  to  differ  from  equation  (33)  in  that  the  term 
(Fi/Fi)  is  not  multiplied  by  c,2,  but  both  (33)  and  (35)  could  be 
made  to  yield  the  same  numerical  value  by  using  a  slightly  different 
value  of  cv  for  the  two.  Custom  has  based  values  of  cv  upon  (35) 
for  the  Venturi  meter  and  upon  (33)  for  the  nozzle. 

With  the  Venturi  meter  we  desire  q,  not  F2,  and  hence,  multi- 
plying (35)  by  Fz  and  replacing  cv  by  c,  we  have 


(36) 


For  a  given  meter  FI  and  F2  are  known  quantities  and,  if  K1  = 
F2\/20/\/(l  —  (Fz/Fi)2,  this  may  be  reduced  to 

q  =  cK'Vh  (37) 

The  coefficient  c  may  be  assumed  to  be  0.985  for  a  new  meter  and 
0.980  for  an  old  one,  the  interior  of  which  will  be  slightly  rougher 
and  perhaps  reduced  in  area  through  incrustation.  These 
factors  will  give  a  result  that  is  very  accurate.  The  coefficient  is 
practically  a  constant,  though  there  is  some  slight  reason  to 


76 


HYDRAULICS 


believe  that  it  increases  slightly  with  higher  rates  of  discharge. 
If  c  is  assumed  constant  for  any  given  meter,  it  is  convenient 
to  replace  cK'  by  K  and  we  then  have 

q  =  K\/h  (38) 

The  Venturi  meter,  invented  by  Clemens  Herschel  in  1886, 
affords  a  most  valuable  and  accurate  means  of  measuring  water, 
especially  in  large  quantities.1  By  a  suitable  recording  device 


Courtesy  of  Builder's  Iron  Foundry. 

FIG.  77. — Venturi  meter  in  wood  pipe  line. 

it  is  possible  to  make  a  continuous  record  of  the  flow  of  water 
through  any  pipe  line  in  which  a  Venturi  meter  is  installed.  The 
sole  objection  to  its  permanent  use  in  a  pipe  is  that  it  must  neces- 
sarily cause  some  slight  friction  loss  or  resistance  to  flow.  If  this 
loss  be  expressed  as  Hf  =  kV22/2g,  we  find  that  values  of  k  range 
from  about  0.1  to  0.2.  The  higher  values  of  k  naturally  go  with 
smaller  values  of  Fz/Fi. 

The  usual  ratio  of  the  diameter  of  the  throat  to  the  diameter  of 

1  Trans.  A.  S.  C.  E.,  vol.  17,  page  228  (1887). 


APPLICATIONS  OF  HYDROKINETICS  77 

the  pipe  is  about  1  to  3,  making  the  ratio  Fz/Fi  =  1/9.  But  in 
order  to  reduce  the  resistance  as  much  as  possible  and  also  to  avoid 
producing  pressures  at  the  throat  below  atmospheric,  it  is  quite 
common  to  make  the  diameters  in  the  ratio  of  1  to  2,  making 
Fz/Fi  =  1/4.  Of  course  this  reduces  the  magnitude  of  h  for  a 
given  rate  of  discharge  and  hence  makes  the  readings  less  accu- 
rate, especially  for  very  low  discharges. 


Courtesy  of  Builder's  Iron  Foundry. 

FIG.  78. — Venturi  meter  of  riveted  steel. 

A  diverging  stream  is  always  less  stable  than  a  converging 
stream,  that  is  it  is  more  readily  broken  up  into  whirlpools  and 
eddies,  and  hence  more  loss  of  energy  takes  place  in  the  portion 
of  the  meter  on  the  downstream  side  of  the  throat.  In  order  to 
minimize  this  the  down  stream  portion  is  made  to  taper  much 
more  gradually  than  the  upstream  side. 

EXAMPLES 

1.  A  Venturi  meter  with  a  4-in.  throat  is  to  be  used  in  a  12-in.  pipe  line. 
Assuming  a  value  of  c  =  0.985,  determine  the  value  of  K  for  this  meter. 

2.  If  a  differential  manometer  employing  mercury  (sp.  gr.  =  13.57)  were 
to  be  used,  determine  the  value  of  K  for  the  Venturi  meter  in  problem  (1), 
replacing  h  by  y  (Fig.  13)  in  inches  of  mercury. 

3.  Suppose  the  throat  of  the  meter  in  problem  (1)  were  to  be  6  in.  the  pipe 
remaining  12  in.     Compute  the  value  of  K. 

4.  Suppose  that  5  cu.  ft.  per  sec.  is  flowing  through  the  Venturi  meter. 
What  are  the  values  of  h  in  problems  (1)  and  (3)? 


78 


HYDRAULICS 


6.  Water  flows  through  a  pipe  line  6  ft.  in  diameter  with  a  velocity  of  7  ft. 
per  sec.  In  this  pipe  line  is  installed  a  Venturi  meter  with  a  throat  diameter 
of  2  ft.  Assuming  the  value  of  k  to  be  0.12,  what  will  be  the  loss  of  head 
caused  by  the  meter?  What  will  be  the  power  lost? 

60.  Large  Vertical  Orifice. — In  the  case  of  an  orifice  whose 
vertical  dimensions  are  large  as  compared  with  its  depth  below 
the  free  surface  it  is  necessary  to  proceed  as  follows :  Choose  an 
elementary  area  dF  such  that  all  portions  are  at  the  same  depth 
z  below  the  free  surface.  Now  by  Art.  52  the  rate  of  discharge 
through  this  strip  may  be  expressed  as 

dq  =  c\/2g~z  dF  (39) 

The  rate  of  discharge  through  the  entire  orifice  may  be  obtained 
by  integrating  equation  (39).     Thus 

q  =  cV2gf^dF  (40) 


T-/-N 

i 
h 

\ 

M 

K                               hz 

f        _       i 

1  \^J 

(I 

FIG.  79. 

(a)  Circular  Orifice 

))   Rectangular  drifice 

In  the  case  of  a  rectangular  orifice  (Fig.  79)  we  may  take 
dF  =  bdz  and  after  integrating  obtain 

q  =  c%V2gb(h^  -hfi  (41) 

If  h  =  depth  of  center  of  gravity  below  the  free  surface  we  may 
write 

h2  =  h  +  d/2-}  hi  =  h  -  d/2. 

Expanding  (h  -f  d/2)^  and  (h  -  d/2)3/i  by  the  binomial  theorem 
and  substituting  in  equation  (41)  we  obtain 


q  =  cbd\/2gh\l  -  gg^  - 


2048/i 


(42) 


The  expression  in  brackets  is  a  rapidly  converging  series  and 
its  value  is  always  less  than  unity.  When  h  =  d,  the  value  of  this 
factor  is  0.989,  while  for  h  =  2d}  its  value  becomes  0.997.  l 
Thus  for  any  head  greater  than  2d  the  rate  of  discharge  may  be 
obtained  by  the  simpler  formula  q  =  cF\/2gh. 
1  Russell,  "Text-book  on  Hydraulics,"  page  60. 


APPLICATIONS  OF  HYDROKINETICS 


79 


In  similar  manner  the  rate  of  discharge  through  a  circular 
orifice  of  area  F  is  given  by  the  expression 

,  _  r  d2  5d4  -| 

q  =  cFVWl  -          2-  ~  '    '    '  J 


It  may  be  found  that  when  h  =  2d  the  value  of  this  series  is 
0.998,  thus  indicating  that  the  use  of  the  exact  formula  is  unnec- 


Free  Surface 


< V— 


\ 


FIG.  80. 

essary  for  heads  above  that  value.  In  Tables  I  and  II  the  coeffi- 
cients in  black-face  type  are  to  be  used  in  the  exact  formulas,  all 
other  coefficients  to  be  used  in  the  formulas  of  Art.  52. 

61.  Weir. — A  weir  is  a  special  form  of  orifice,  its  distinguishing 
feature  being  that  it  is  placed  at  the  water  surface  so  that  the 
head  on  its  upper  edge  is  zero.  Thus  the  usual  formulas  for  ori- 


FIG.  81.— Weir. 

fices  given  in  Art.  52  can  no  longer  be  applied  and  the  methods  of 
Art.  60  must  be  employed.  The  weir  is  one  of  the  most  widely 
accepted  standard  devices  for  the  measurement  of  water. 

If  it  be  assumed  that  the  velocity  of  water  through  an  orifice 
varies  as  the  square  root  of  the  depth,  the  curve  in  Fig.  80  would 
give  a  true  representation  of  the  flow.  However,  the  particles  of 


80 


HYDRAULICS 


water  at  the  surface  of  the  weir  opening  do  not  remain  at  rest  but 
flow  "with  considerable  velocity.  It  may  also  be  observed  that 
the  level  of  the  water  at  this  point  drops  below  its  normal  value,  as 
shown  in  Fig.  81.  Also  it  must  be  noted  that  the  stream  lines 

flowing  through  the  weir  are  not  neces- 
sarily normal  to  the  plane  of  the  weir; 
hence  it  is  hardly  correct  to  multiply 
their  velocities  by  areas  in  the  plane 
of  the  latter.  For  these  and  other 
reasons  it  is  impossible  to  derive  by 
theory  weir  formulas  which  are  ex- 
actly correct,  but  they  serve  as  ex- 
pressions which  may  be  made  to  yield 
correct  results  by  the  proper  choice 
and  use  of  experimental  coefficients. 
It  might  seem  natural  to  measure 
the  depth  of  water  flowing  over  the 
crest  of  a  weir,  but  in  practice  it  is 
difficult  to  do  this  with  any  degree  of 
accuracy.  It  is  found  more  feasible 
to  measure  the  elevation,  above  the 
weir  crest,  of  the  water  surface  at 
some  distance  back  from  the  weir, 
where  the  water  is  relatively  quiet. 
Thus  all  weir  formulas  express  the  rate 
of  discharge  as  a  function  of  H  (Fig. 
81).  This  measurement  must  be 
taken  at  a  point  far  enough  back  to 
avoid  the  effects  of  the  surface  curve. 
This  distance  should  be  at  least  QH. 
The  usual  instrument  for  measuring 
H  is  the  hook  gage,  one  form  of 

w.  &  L.  E.  Guriey.  which  is  shown  in  Fig.  82.     The  gage 

FIG.  82. — Hook  gage.  should  be  mounted  on  some  rigid  sup- 
port. In  using  it  the  sharp-pointed  hook  is  submerged  beneath 
the  surface  and  then  carefully  raised  until  a  slight  distortion  may 
be  seen  on  the  water  surface.  The  hook  should  then  be  lowered 
until  this  distortion  barely  disappears.  From  this  reading  the 
value  of  H  is  obtained  by  subtracting  the  "-hook  gage  zero," 
which  is  the  reading  of  the  gage  when  its  point  is  just  level  with 
the  crest  of  the  weir,  as  the  lower  edge  is  called. 


APPLICATIONS  OF  HYDROKINETICS 


81 


The  Triangular  Weir. — The  triangular  weir  such  as  is  shown  in 
Fig.  83  is  useful  for  measuring  relatively  small  rates  of  discharge, 
as  a  reasonable  value  of  H  may  be  obtained  by  employing  a 
sufficiently  small  vertex  angle.  But  for  discharges  much  above 
2  or  3  cu.  ft.  per  sec.  excessively  high  values  of  H  are  necessary 
and  other  types  of  weir  would  then  be  used. 

The  Suppressed  Rectangular  Weir. — Probably  the  most  common 
type  of  weir  is  one  whose  shape  is  rectangular.  If  the  width  of 


From  a  photograph  by  the  author. 
FIG.  83. — Discharge  from  a  60°  triangular  weir. 

the  weir  is  the  same  as  that  of  the  channel  .of  approach,  as  in 
Figs.  84  and  85,  the  stream  of  water  flowing  over  the  crest  will 
not  undergo  any  lateral  contraction,  that  is  the  end  contractions 
are  suppressed.  With  this  type  of  weir  it  is  customary  to  extend 
the  sides  of  the  channel  beyond  the  crest  so  that  the  falling  stream 
is  bounded  by  them.  If  these  sides  are  not  so  extended  the 
stream  will  expand  somewhat  and  the  discharge  for  a  given  value 
of  H  will  be  slightly  larger  than  in  the  standard  type. 

It  is  necessary  in  this  or  any  type  of  weir  to  insure  that  the  weir 
is  "  ventilated,"  that  is  that  air  has  access  to  the  under  side  of  the 


82 


HYDRAULICS 


falling  water.  Otherwise  the  air  will  be  gradually  swept  out  and 
the  water  will  tend  to  cling  to  the  face  of  the  weir  instead  of 
springing  clear  of  it.  For  a  given  value  of  H  the  rate  of  discharge 


From  a  photograph  by  the  author. 

FIG.  84. — Rectangular  weir  without  end  contractions. 


From  a  photograph  by  the  author. 

FIG.  85. — Rectangular  weir  without  end  contractions. 

would  then  be  greatly  increased  and  the  usual  coefficients  would 
no  longer  apply. 

In  order  to  insure  that  the  water  shall  spring  clear,  that  is  that 


APPLICATIONS  OF  HYDROKINETICS 


83 


perfect  crest  contraction  shall  be  attained,  it  is  necessary  to  have 
a  sharp  edge  on  the  weir  plate.  This  may  be  produced  by  bevel- 
ing the  edge  of  a  metal  plate  down  to  a  knife  edge.  However,  a 


From  a  photograph  by  F.  H.  Fowler. 

FIG.  86. — Rectangular  weir  with  end  contractions. 


From  a  photograph  by  the  author. 

FIG.  87. — Trapezoidal  or  Cippoletti  weir. 

perfectly  sharp  square  shoulder  is  just  as  good  as  a  knife  edge  if 
the  plate  is  not  too  thick  for  the  water  to  clear  the  other  shoulder. 
For  very  low  values  of  H  the  knife  edge  would  still  permit  crest 
contraction  to  take  place  when  the  flat  edge  of  a  plate  would  not. 


84  HYDRAULICS 

Contracted  Rectangular  Weir. — When  the  width  of  the  weir  is 
less  than  that  of  the  channel  of  approach,  as  in  Fig.  86,  the 
contraction  that  occurs  at  each  end  causes  the  real  width  of  the 
stream  of  water  to  be  less  than  that  of  the  weir  itself.  Such  a 
weir  is  called  a  contracted  weir. 

Trapezoidal  or  Cippoletti  Weir. — The  trapezoidal  weir  is  one 
in  which  the  sides  of  the  notch  are  not  vertical  but  diverge  so 
that  the  width  at  the  water  surface  increases.  If  the  side  slopes 
have  the  ratio  of  1:4  the  weir  is  called  a  Cippoletti  after  an 
Italian  engineer  of  that  name  who  proposed  it.  The  advantage 
of  this  type  of  weir  will  be  stated  later. 


FIG.  88. 


62.  The  Triangular  Weir.  —  In  Fig.  88  we  have  a  triangular 
weir  with  any  vertex  angle  6.  The  rate  of  discharge  through  an 
elementary  strip  of  area  dF  will  be 


dq  =  cV2gz  dF.  ' 

Now  dF  =  xdz  and  by  similar  triangles  x  :  b  =  (H  —  z) :  H . 
Hence  dF  =  jj  (H  —  z)dz.  Substituting  in  the  above  we  have 
for  the  entire  notch 

Q  = 
Integrating  between  limits  we  have 

q  =  cV2g  jj  [%H*  -  %H»]. 
But  b  =  2H  tan  6/2.     Inserting  this  and  reducing  we  have 

(44) 


APPLICATIONS  OF  HYDROKINETICS 
This  expression  for  any  given  weir  may  be  reduced  to 


85 


q  =  KH  (45) 

In  Figs.  89  and  90  may  be  found  experimental  values  of  K  for  sev- 
eral triangular  weirs.  The  54°,  60°,  and  one  of  the  90°  weirs  are 
in  the  laboratory  of  Sibley  College.1  The  two  lower  curves  in 
Fig.  90  were  plotted  from  data  in  a  very  valuable  paper  by  James 
Barr.2  The  weir  for  which  the  very  lowest  curve  was  constructed 


1.60 


1.50 


1.40 


1.30 


1.20 


60  Triangular  Weir 


54  Triangular 


Weir 


0.1  0.2 

FIG.  89. — Coefficients  of  triangular  weirs. 


0.3  0.4  0.5  0.6  0.7 

H  in  Feet 


0.8 


2.60 

t 

§2.50 

2.40 
( 

7 

—  . 
•'  •••••  •— 

^^ 

=^ 

\ 

90 

Tris 

ngu 

ar  V 

reir 

7 

^r: 

==: 

'  _ 

^          — 

, 

=T- 

• 

•^"  

/ 

)              0.1            0,2            0.3            0.4            0.5            0.6             0.7            0.8 

H  in  Feet 
FIG.  90.— Coefficients  of  90°  triangular  weirs. 

had  a  very  fine  sharp  edge,  while  the  other  weir  had  a  square 
corner  and  a  thickness  of  about  He  m-  Both  of  these  weirs  have 
values  of  K  below  that  of  the  Sibley  College  weir  but  the  differ- 
ence, of  about  1  per  cent.,  may  be  due  to  different  methods  of 
measurement  of  the  water  and  of  H. 

1  Engineering  News,  vol.  73,  page  636  (1915). 

2  "Experiments  upon   the   Flow   of   Water   over   Triangular   Notches." 
Engineering,  Apr.  8  and  15,  1910. 


86  HYDRAULICS 

These  curves  show  either  that  the  discharge  does  not  vary  as 
the  five-halves  power  of  H  or  that  c  is  not  a  constant.  Thomp- 
son, who  first  employed  the  triangular  weir,  chose  for  K  a  value 
of  2.54.  The  value  of  2.65  that  so  many  writers  persist  in  giving 
is  entirely  too  high  and  is  based  upon  Thompson's  first  experi- 
ments which  were  not  accurately  performed. 

(In  finding  H5/i  on  some  slide  rules,  it  is  well  to  note  that 
H*  =  H*V3.) 

63.  The  Rectangular  Weir. — For  the  rectangular  weir  in  Fig. 
91  the  discharge  through  the  elementary  strip  of  area  dF  may  be 
given  by 

dq  =  c\/2gz  bdz. 


-     H 

— L 


FIG.  91. 

Integrating  between  limits  we  have 

CH 

q  =  c\/2g  b  I  e*  dz 

q  =  \  cV2g  bH^  (46) 

This  is  the  fundamental  formula.  Many  variations  of  it  have 
been  suggested  in  an  attempt  to  express  the  value  of  c,  which  is 
not  necessarily  a  constant  for  all  values  of  H.  (It  may  be  noted 
that  H*  =  H^.) 

64.  The  Francis  Weir  Formula.— About  1850  Mr.  James  B. 
Francis  made  some  very  accurate  investigations  of  the  flow  of 
water  over  weirs.1  As  a  result  of  his  experiments  he  selected  a 
value  of  0.622  for  c  in  equation  (46),  so  that  for  a  suppressed  weir 
we  have 

q  =  3.33bHH  (47) 

With  a  contracted  weir  he  concluded  that  the  effect  of  each  con- 
traction was  to  reduce  the  effective  width  of  the  weir  by  Q.IH. 
Thus  for  a  contracted  weir  we  have 

q  =  3.33(6  -  0.1  nH)HH  (48) 

1  "Lowell  Hydraulic  Experiments." 


APPLICATIONS  OF  HYDROKINETICS  87 

The  usual  type  of  contracted  weir  will  have  two  end  contractions, 
giving  n  a  value  of  2  in  equation  (48),  but  we  might  have  a  weir 
with  one  end  contracted  and  one  end  without  contraction. 
Equation  (48)  is  strictly  empirical  and  applicable  only  within 
limits.  If  H  is  greater  than  one-third  b  it  is  impossible  for  perfect 
end  contraction  to  occur  and  hence  the  conditions  upon  which 
(48)  is  based  no  longer  exist. 

When  the  cross-section  area  of  the  channel  of  approach  is  rela- 
tively small,  there  may  be  a  velocity  of  flow  in  it  that  is  high 
enough  to  affect  the  result.  This  velocity  is  called  the  velocity 
of  approach.  Francis  corrected  for  this  by  replacing  H^  in 
both  (47)  and  (48)  by  [(H  +  hv}^  -  h^]}  where  hv  is  the  velocity 
head  in  the  channel.  In  practical  work  the  last  term  is  often 
dropped.  There  is  no  real  good  theoretical  foundation  for  any 
expression  involving  velocity  of  approach.  Others  replace  the 
H^  in  the  formulas  by 

(H  +  ahj* 

in  which  a  is  given  values  ranging  from  1  to  2.  If  the  velocity 
of  water  in  the  channel  at  the  section  where  H  is  measured  is 
uniform  over  the  cross-section,  a  value  of  1  should  be  used  for  a. 
But  if  the  surface  velocity  is  much  higher  than  the  bottom  veloc- 
ity, the  value  of  a.  should  be  greater  than  unity.  This  is  because 
the  true  velocity  head  which  affects  the  discharge  over  the  weir  is 
greater  than  the  hv  which  is  based  upon  the  average  velocity  in 
the  section.  In  correcting  for  velocity  of  approach  the  first  step 
would  be  to  solve  for  an  approximate  value  of  q  by  either  equation 
(47)  or  (48)  as  the  case  might  be.  Then  this  value  of  q  should  be 
divided  by  the  area  of  the  channel  of  approach  at  the  section 
where  H  is  measured.  From  this  velocity  an  approximate  value 
of  hv  can  be  obtained.  This  would  be  added  to  H  and  a  new 
and  somewhat  larger  value  of  q  calculated.  From  this  a  new 
value  of  hv  could  be  computed,  and  so  on.  However,  after  about 
two  such  solutions  it  would  be  found  that  further  solution  would 
alter  the  result  very  slightly. 

65.  The  Bazin  Weir  Formula. — Bazin,  in  France,  made  a  valu- 
able series  of  experiments  upon  weirs  without  end  contractions 
and  with  high  velocities  of  approach.  From  these  he  devised  a 
weir  formula  which  expresses  the  effect  of  velocity  of  approach  in  a 
much  less  awkward  manner  than  the  Francis  formula.  His  most 


88 


HYDRAULICS 


accurate  formula  is  rather  complicated,  but  for  practical  work  the 
following  approximate  formula  is  sufficient: 

J  \     9T 

(49) 


The  quantity  y  indicates  the  height  of  the  weir  crest  above  the 
bottom  of  the  channel  and  thus  introduces  the  effect  of  velocity 
of  approach  into  the  formula  in  an  indirect  manner. 

66.  Comments  on  Weirs. — The  formula  (48)  for  the  contracted 
weir  is  applicable  only  to  the  standard  Francis  weir,  whose  lim- 
iting proportions  are  shown  in  Fig.  92.  Unless  there  is  sufficient 
space  left  at  the  two  ends  and  at  the  bottom  of  the  weir,  perfect 
contraction  will  not  be  obtained.  These  dimensions  may  be 


—  2H 


—  2H 


I 
2H 


FIG.  92. — Limiting  proportions  of  standard  contracted  weirs. 

made  greater  than  the  values  given,  but  never  less.  The  height 
of  the  crest  above  the  bottom  should  preferably  be  at  least  3H. 
Francis  estimated  that  a  height  of  2H  would  increase  the  dis- 
charge about  one  per  cent.  If  sufficient  space  cannot  be  had 
to  secure  perfect  end  contraction,  the  end  contractions  should 
be  entirely  suppressed,  or  one  of  them  suppressed  and  all  the 
space  given  to  the  other  end.  There  is  no  known  coefficient  or 
method  for  dealing  with  imperfect  contraction.  Also  the  value 
of  H  should  not  exceed  one-third  of  the  width  b. 

For  the  suppressed  weir  there  are  no  standard  dimensions  to  be 
observed.  Experiments  at  Cornell  University  have  indicated 
that  the  Francis  coefficient  in  equation  (47)  is  applicable  for  heads 
up  to  at  least  5  ft.  and  for  heads  down  to  0.3  ft.  For  H  =  0.2 
ft.  it  should  be  increased  3  per  cent.,  and  for  H  =  0.1  ft.  it  should 
be  increased  7  per  cent. 

If  the  area  of  the  channel  of  approach  exceeds  QbH  it  can  be 
shown  that  the  velocity  of  approach  is  negligible.  Hence  it  is 


APPLICATIONS  OF  HYDROKINETICS  89 

seen  that  velocity  of  approach  need  not  be  considered  in  a  con- 
tracted rectangular  weir.  But  with  a  suppressed  weir  the  depth 
of  the  channel  would  have  to  be  6H,  and  that  is  not  often  the  case. 
Hence  most  suppressed  weirs  have  a  velocity  of  approach  that 
needs  to  be  considered.  The  Francis  coefficient  was  based  upon 
work  with  weirs  having  a  velocity  of  approach  less  than  1  ft.  per 
sec.  When  the  velocity  of  approach  is  high,  the  formula  of 
Bazin  should  be  applied. 

The  most  accurate  type  of  weir  is  a  suppressed  weir  with  such 
a  deep  channel  of  approach  that  the  velocity  of  approach  is  neg- 
ligible. A  contracted  weir  for  which  the  velocity  of  approach  is 
negligible  is  about  in  the  same  class  with  a  suppressed  weir  with 
a  moderate  velocity  of  approach.  End  contractions  have  been 
held  to  be  a  source  of  error  and  there  appears  to  be  no  truly 
rational  way  to  correct  for  them.  The  least  desirable  type  of 
weir  is  the  one  with  a  high  velocity  of  approach  because  of  the 
difficulty  not  only  of  reading  H  accurately  but  also  of  allowing 
for  the  effect  of  this  velocity  in  a  scientific  manner. 

It  almost  goes  without  saying  that  a  weir  should  be  set  with  its 
crest  level  and  its  plane  vertical.  An  inclination  upstream 
decreases  and  an  inclination  downstream  increases  the  dis- 
charge for  a  given  H.  The  crest  should  be  sharp  and  in  good 
condition. 

In  using  weirs  for  accurate  work  it  is  desirable  to  study  the 
original  experiments  upon  which  the  formulas  are  based  and  use 
the  formula  that  has  been  derived  under  circumstances  most 
nearly  like  those  in  hand.  And  it  is  likewise  desirable  to  dupli- 
cate the  original  investigator's  methods.  The  hook  gage,  for 
instance,  should  be  located  in  the  same  way  and  at  the  same 
distance  from  the  weir.  Unless  these  precautions  are  followed 
one  has  no  assurance  that  the  coefficients  given  fit  his  own 


case. 


67.  The  Cippoletti  Weir. — In  order  to  avoid  the  trouble  of 
correcting  for  end  contractions,  the  sides  of  the  Cippoletti  weir 
are  given  such  a  batter  (1:4)  that  they  add  enough  to  the  effect- 
ive width  of  the  stream  to  offset  the  contraction  Q.2H  of  the  con- 

1  For  information  on  weirs  see: 

"Weir  Experiments,  Coefficients,  and  Formulas,"    by    R.    E.    Horton, 

U.  S.  G.  S.  Water  Supply  and  Irrigation  Paper,  No.  150,  Revised  No.  200. 

Hughes  and  Safford,  "Hydraulics." 

Parker,  "The  Control  of  Water." 


90 


HYDRAULICS 


tracted  Francis  weir.     Thus  computations  may  be  made  upon 
the  basis  of  the  width  6  at  the  crest  by  the  following  formula 

q  =  3.3676H94  (50) 

68.  Special  Weirs. — There  are  other  types  of  weirs  for  special 
purposes.  Thus  we  have  a  floating  circular  intake  weir  in  Fig. 
93.  The  purpose  of  this  is  not  for  accurate  water  measurement, 
but  rather  to  float  in  such  a  manner  in  the  reservoir  that  a  certain 


From  a  photograph  by  F.  H.  Fowler. 

FIG.  93. — Floating  circular  intake  weir  for  Los  Angeles  aqueduct. 

depth  of  water  continually  flows  through  or  over  its  passageways 
and  down  the  intake  in  the  center.  Thus  no  matter  how  the 
water  level  in  the  reservoir  rises  and  falls  a  uniform  quantity  will 
automatically  be  delivered  to  the  intake. 

69.  The  Pitot  Tube. — Among  other  water-measuring  devices 
is  the  Pitot  tube.  This  is  an  instrument  which  indicates  the 
velocity  of  water  at  a  point.  From  the  velocity  the  rate  of  dis- 
charge may  be  obtained. 

The  principle  of  the  Pitot  tube  is  illustrated  in  Fig.  94  and  its 
theory  will  be  discussed  in  a  subsequent  chapter.  For  an  open 
stream  only  a  single  tube  is  necessary,  but  in  a  stream  of  water 
under  pressure  a  second  tube  is  necessary  to  record  the  pressure 


APPLICATIONS  OF  HYDROKINETICS 


91 


alone.  The  quantity  desired  is  the  difference  between  the  two 
readings,  which  we  shall  call  h.  It  can  be  shown  by  correct 
theory  that  if  h  is  the  value  in  feet  of  water  of  the  dynamic  pres- 
sure exerted  by  the  impact  of  the  stream  against  the  opening  of 
the  tube,  the  velocity  of  the  water  is  given  by 

V  =  V5gh  (51) 


_i 


FIG.  94. 

This  has  been  found  by  experiment  to  be  true  when  there  is 
smooth  stream  line  flow,  but  in  case  of  turbulent  flow  we  should 
introduce  a  coefficient  whose  value  is  about  0.977,  so  that  we 
should  write 

V  =  0.977\/20£  (52) 

The  fact  that  this  coefficient  is  anything  less  than  unity  is  not 
because  our  theory  is  at  fault  nor  because  of  any  defect  in  the  in- 
strument itself,  but  is  due  to  the  fact  that  the  instrument  records 
the  true  velocity  at  the  point  while  we  desire,  for  practical  pur- 
poses, the  axial  component  of  velocity.  Hence  the  factor  is 
designed  to  give  us  the  axial  component  of  velocity  (OB  in  Fig. 
52  rather  than  OD).1 

In  using  the  Pitot  tube  it  is  often  convenient  to  divide  a  cross- 
section  of  the  stream  up  into  parts  of  equal  area  and  to  determine 
the  velocity  in  the  center  of  each  area.  The  average  velocity  of 

1 L.  F.  Moody,  "Measurement  of  the  Velocity  of  Flowing  Water."  Proc. 
of  the  Engineer's  Soc.  of  W.  Penn.,  vol.  30,  page  319  (1914). 


92  HYDRAULICS 

the  stream  will  be  the  average  of  the  observed  velocities.  But  if 
the  areas  are  not  equal  the  average  of  the  velocities  will  have  no 
significance.  It  will  then  be  necessary  either  to  plot  a  curve 
from  which  velocities  at  other  points  may  be  taken  or  to  multiply 
each  observed  velocity  by  the  area  which  it  may  be  assumed  to 
represent.  The  total  rate  of  discharge  of  the  entire  stream  is  the 
sum  of  all  such  partial  discharges.  Thus 

q  =  2F'V  (53) 

where  F'  is  a  portion  of  the  total  area  and  V  is  the  velocity 
through  that  area.  If  the  average  velocity  is  desired,  it  can  be 
obtained  by  dividing  the  rate  of  discharge  by  the  total  area. 

70.  The  Current  Meter. — For  moderate  velocities  such  as  are 
found  in  canals  and  natural  streams  the  current  meter  is  well 
adapted.     It  consists  of  a  wheel,  as  in  Fig.  95,  or  in  other  types  a 
screw,  which  is  rotated  by  the  action  of  the  water.     By  calibra- 
tion the  relation  is  determined  between  the  velocity  of  the  water 
and  the  rate  of  rotation  of  the  meter. 

In  many  current  meters  each  revolution  is  recorded  by  a  click 
in  a  telephone  receiver  at  the  ear  of  the  observer,  the  click  being 
produced  by  the  wheel  making  an  electric  contact  every  revolu- 
tion. In  some  meters  the  contact  is  not  made  so  frequently. 

In  using  the  current  meter  the  velocities  are  determined  at  a 
number  of  different  points  and  the  total  discharge  of  the  entire 
stream  computed  in  the  same  manner  as  in  Art.  69. 

71.  Comments   on   Measurement   of   Water. — The  accurate 
measurement  of  rate  of  discharge  is  one  of  the  most  difficult 
problems  in  practical  hydraulics.     The  only  positive   way  of 
measuring  rate  of  discharge  is  to  weigh  the  amount  of  water  dis- 
charged in  a  given  time  or  to  determine  its  volume  in  suitably 
calibrated  tanks  or  reservoirs.     The  former  method  is  applicable 
only  for  relatively  small  rates  of  discharge,  and  facilities  for  the 
latter  are  seldom  to  be  had.     Also  in  the  latter  method  the  effect 
of  leakage,  evaporation,  and  other  factors  may  sometimes  prove 
troublesome. 

The  methods  that  are  usually  employed  are  the  ones  that  have 
been  given  in  this  chapter.  They  are  all  indirect  in  that  we  as- 
sume the  velocity  or  the  rate  of  discharge  to  be  a  function  of  some 
other  quantity  which  can  be  measured. 

The  discharge  of  water  from  any  tank  can  be  measured  by  an 
orifice,  tube,  or  nozzle.  When  a  stream  of  water  flows  in  an  open 


APPLICATIONS  OF  HYDROKINETICS 


93 


channel  it  may  be  caused  to  flow  over  a  weir  or  its  velocity  through- 
out any  cross-section  may  be  found  by  a  current  meter,  by  floats, 
or  other  means.  For  a  stream  of  water  confined  within  a  closed 
pipe  we  may  use  a  Pitot  tube  to  determine  the  velocity  across 
a  cross-section  or  cause  the  water  to  flow  through  a  Venturi 


Courtesy  of  W.  &  L.  E.  Curley. 

FIG.  95.- 


-Current  meter. 


meter.  At  the  end  of  a  pipe  line  we  might  place  a  nozzle  which 
would  also  permit  the  rate  of  discharge  to  be  obtained.  The 
discharge  from  a  nozzle  may  be  computed  or  it  may  be  measured 
directly  by  determining  the  velocity  of  the  jet  with  a  Pitot  tube. 
The  means  of  measurement  that  is  to  be  used  depends  upon 
the  circumstances. 


94  HYDRAULICS 

In  addition  to  the  methods  of  measurement  that  have  been 
described  in  this  chapter,  there  are  other  methods,  especially 
chemical  methods.  One  of  these  is  simply  a  matter  of  discharg- 
ing a  small  quantity  of  highly  colored  liquid  into  the  intake  of  a 
pipe  line  and  noting  the  time  that  it  takes  for  the  discoloration  to 
be  noted  at  the  other  end.  Knowing  the  length  of  the  pipe  it  is 
easy  to  compute  the  velocity  of  the  water.  Another  valuable 
method  consists  of  adding  a  strong  salt  solution  and  by  analysis 
of  the  water  taken  at  a  downstream  section  the  amount  of  dilu- 
tion is  ascertained  and  hence  the  rate  of  discharge  of  the  stream 
may  be  determined. 

72.  Discharge  under  Varying  Head. — If  the  head  varies,  the 
rate  of  discharge  will  likewise  vary  and  the  total  discharge  in  a 
given  time,  or  the  time  required  for  a  given  total  discharge,  must 
be  determined  as  follows: 

Let  the  head  on  any  orifice  be  z,  a  variable  quantity.     Then 

q  =  cF\S2gz. 

And  in  any  time  dt  we  should  have  a  total  amount  of  water  qdt 
discharged,  or 

qdt  =  cF\f2gz  dt. 

In  the  time  dt  suppose  the  water  surface  is  lowered  an  amount  dz. 
If  the  area  of  the  water  surface  be  denoted  by  A  then 

qdt  =  -Adz 

the  minus  sign  being  used  because  z  decreases  as  t  increases. 
Placing  this   value  of  qdt  in  the  above  expression   we  have 

-Adz  =  cF\^2gzdt 
This  may  be  written  as 

cFV2g  dt  =  -A-~  (54) 

In  the  above  equation  A  may  be  a  constant  or  it  may  be  some 
function  of  z.  In  either  event  the  integration  of  the  equation 
will  enable  the  desired  results  to  be  obtained. 

73.  PROBLEMS 

1.  What  is  the  rate  of  discharge  of  a  54°  triangular  weir  when  H  =  0.400 
ft.?  With  the  same  value  of  H  what  would  be  the  rate  of  discharge  of  a  90° 
triangular  weir? 


APPLICATIONS  OF  HYDROKINETICS  95 

2.  What  would  be  the  value  of  H  for  a  rate  of  discharge  of  2.0  cu.  ft.  per 
sec.,  if  a  60°  triangular  weir  is  used?     Assume  K  =  1.45. 

3.  The  width  of  the  weir  in  Fig.  84  is  7.573  ft.     Neglecting  velocity  of 
approach,  what  is  the  rate  of  discharge  when  H  =  1.200  ft.? 

4.  In  problem  (3),  if  the  value  of  y  is  2.85  ft.,  solve  for  the  rate  of  discharge 
by  using  the  Francis  formula  with  a  =  1.0.     Solve  with  a  =  2.0. 

6.  Solve  problem  (4)  by  the  use  of  Bazin's  formula. 

6.  Assume  that  the  weir  in  Fig.  86  is  also  7.573  ft.  in  width.     What  would 
be  the  rate  of  discharge  when  H  =  1.200  ft.? 

7.  Assume  that  a  Pitot  tube  and  a  piezometer  tube  are  connected  to  two 
sides  of  a  differential  manometer  containing  mercury.     Suppose  the  Pitot 
tube  is  placed  in  such  positions  in  the  stream,  which  is  10  in.  in  diameter,  that 
it  measures  the   velocities  in   five  areas  of   equal  magnitude.     Suppose 
these  five  readings  on  the  differential  manometer  are  1.50,  2.15,  2.84,  3.62, 
and  4.05  in.  of  mercury.      Find  the  rate  of  discharge  of  the  stream. 

8.  Suppose  that  a  ship  lock  in  a  canal  has  a  uniform  area  of  water  surface 
at  all  depths  of  water,  what  would  be  the  integration  of  equation  (54)  ? 

9.  Suppose  that  a  ship  lock  in  a  canal  is  of  uniform  rectangular  cross-sec- 
tion and  that  it  is  300  ft.  by  90  ft.  by  40  ft.  deep.     Suppose  that  the  water 
from  this  lock  is  discharged  through  a  tunnel  which  is  3  ft.  in  diameter,  the 
coefficient  of  discharge  being  0.50.     If  the  initial  head  under  which  water 
discharges  is  35  ft.,  how  long  will  it  take  for  the  level  to  drop  25  ft.? 

10.  In  the  problem  (9)  how  large  would  the  tunnel  have  to  be  to  permit  the 
water  level  to  drop  25  ft.  in  15  min.? 


CHAPTER  VII 
,      FLOW  THROUGH  PIPES 

74.  Loss  of  Head  in  Pipe  Friction. — In  dealing  with  such 
devices  as  the  orifice,  nozzle,  Venturi  meter,  etc.,  we  have 
compensated  for  the  effect  of  frictional  resistance  to  flow  by  the 
introduction  of  velocity  coefficients.  This  is  feasible  because 
all  of  these  devices  can  be  standardized  so  that  the  coefficients 
which  have  been  determined  for  one  may  be  applied  to  another 
of  the  same  type.  We  might  have  velocity  coefficients  for  pipes 
also  if  the  latter  were  more  nearly  alike.  But  actually  pipe  lines 
differ  from  each  other  in  length,  size,  degrees  of  roughness,  and 
other  respects  to  such  an  extent  that  the  application  of  velocity 
coefficients  is  impractical.  Therefore  it  is  necessary  to  proceed 
on  a  different  basis. 

In  Art.  44  it  is  stated  that  the  loss  of  head  which  always 
accompanies  flow  may  be  expressed  as 

H'  =  k^  (55) 

i 

In  the  case  of  a  pipe  line,  or  any  water  conduit  of  any  length,  it 
is  apparent  that  the  loss  of  head  between  two  sections  is  a 
function  of  the  distance  between  them  and  that  the  factor  k 
should  be  proportional  to  the  roughness  of  the  surface  of  the 
channel.  Mathematical  analysis  as  well  as  experimental  evi- 
dence indicates  that  the  friction  is  less  in  a  large  conduit  than  in  a 
small  one.  It  is  also  found  that  hydraulic  friction  is  independent 
of  the  pressure  and  the  temperature  effect  is  so  slight  that  it  can 
be  neglected. 

In  order  to  express  fc  as  a  function  of  the  size  of  the  channel  we 
need  some  dimension  which  can  be  used  for  all  shapes  of  cross- 
section.  In  the  case  of  a  circular  pipe  alone  we  might  use  the 
diameter/  but  this  would  not  be  applicable  for  other  shapes. 
The  quantity  that  is  used  for  this  purpose  is  the  ratio 

area  of  water  cross-section  .     . 

length  of  wetted  perimeter 
96 


FLOW  THROUGH  PIPES  97 

This  is  a  linear  dimension  and  is  called  the  "  hydraulic  mean 
depth"  or  the  "hydraulic  radius."  Its  physical  meaning,  so 
far  as  it  has  any,  is  that  m  is  the  depth  of  water  necessary  on  a 
plane  surface  of  width  equal  to  the  length  of  the  wetted  perime- 
ter, so  that  the  imaginary  volume  of  water  thus  formed 
shall  be  equal  to  the  actual  volume. 

Mathematical  analysis  and  experimental  investigation  have 
led  to  an  approximate  empirical  formula 

fc  =  7-  (57) 

4  m 

where  /  is  a  friction  coefficient  which  depends  upon  the  roughness 
of  the  surface  as  well  as  upon  other  factors. 

Experimental  evidence  indicates  that  the  value  of  n  in  equation 
(55)  varies  from  about  1.75  to  2.00,  the  former  holding  for  very 
smooth  surfaces  and  the  latter  for  rough  ones  such  as  the  interiors 
of  iron  and  steel  pipes  after  years  of  service.1  As  shown  in  Art. 
84  it  is  possible  to  select  values  of  n  for  different  kinds  of  surfaces 
but  ordinarily  this  is  not  done  since  it  is  difficult  to  express  degree 
of  roughness  with  any  precision  and  often  the  degree  of  roughness 
cannot  be  known  or  estimated.  Hence  for  the  sake  of  simplicity 
in  computation  the  value  n  =  2.00  is  usually  used. 

Assuming  n  =  2.00  and  inserting  the  value  of  k  given  by  (57) 
in  (55),  we  have, 

ff'^Z-TT  (58) 

4  m  2g 

If  the  channel  is  a  circular  pipe  flowing  full  of  water  the  value 
of  m  is 

m  =  7rr2/27rr  =  r/2  =  d/4. 

Inserting  this  value  of  m  in  equation  (58)  we  have  for  a  circular 
pipe  full  of  water, 

H'-f~d%  (59) 

If  equation  (59)  were  entirely  correct  we  should  expect  that  / 
would  depend  only  upon  the  roughness  of  the  surface.  But  we 
have  already  seen  that  the  loss  of  head  does  not  always  vary  as 
the  square  of  the  velocity.  If  the  actual  exponent  of  V  is  less 
than  2,  the  value  of /in  equation  (59)  would  have  to  decrease 
with  increasing  velocity.  But  for  rough  surfaces,  where  the 

1  For  velocities  below  the  critical  the  loss  of  head  varies  as  the  first  power 
of  the  velocity.     It  is  then  also  a  function  of  temperature. 

7 


98  HYDRAULICS 

exponent  is  practically  2,  the  value  of  /in  (59)  should  be  inde- 
pendent of  the  velocity.  Fortunately  the  variation  of  /  with  V 
is. not  very  great  in  most  cases  of  actual  practice  and  may  be 
neglected  in  the  present  treatment.  It  has  also  been  found  that 
the  loss  of  head  does  not  vary  inversely  as  the  first  power  of  d 
but  rather  as  d1-25.  Thus  if  equation  (59)  is  used,  the  value  of/ 
should  be  made  to  decrease  as  d  increases. 

According  to  the  experiments  of  Darcy  the  value  of  /  for  new, 
clean,  cast-iron  pipes  may  be  given  by 

/  -  0.02  +  ^  (60) 

For  old,  corroded,  cast-iron  pipe  the  values  given  by  (60)  should 
be  doubled,  but  it  is  impossible  lo  formulate  any  definite  law  by 
which  the  value  of  /  should  be  increased  with  age.  It  depends 
to  some  extent  upon  the  chemical  composition  of  the  water 
carried.  In  the  case  of  a  smooth  wood-stave  pipe  the  value  of 
/  should  be  somewhat  less  than  that  given  by  (60)  and  it  does 
not  increase  with  age.  For  a  riveted  steel  pipe  the  values  of  / 
are  slightly  greater  than  given  by  (60).  But  roughness  cannot  be 
expressed  mathematically  and  the  selection  of  /  for  any  given 
case  is  largely  a  matter  of  judgment.1 

It  should  be  noted  that  in  (60)  the  value  of  d"  is  in  inches,  but 
/  is  an  abstract  number.  In  equation  (59)  we  have  another 
abstract  number,  the  ratio  of  the  length  to  the  diameter.  Thus 
I  and  d  should  both  be  in  the  same  units. 

EXAMPLES 

1.  What  will  be  the  loss  of  head  in  a  10-in.  pipe  line  2,000  ft.  long,  when 
the  velocity  of  the  water  in  the  pipe  is  6  ft.  per  sec.  ? 
Ans.  29.5  ft. 


FIG.  96. 

2.  Suppose  that  the  pipe  of  problem  (1)  is  shown  in  Fig.  96,  that  there 
are  no  losses  save  those  due  to  pipe  friction,  and  that  the  size  of  the  stream 
discharging  at  the  end  is  the  same  as  that  of  the  pipe.  What  value  of  h 
will  be  required  to  produce  the  flow  in  problem  (1)?  Ans.  30  ft. 

1  Unless  otherwise  specified,  all  problems  in  this  text  will  be  based  upon 
the  value  of  /  given  by  (60)  for  the  sake  of  uniformity. 


FLOW  THROUGH  PIPES 


99 


3.  Suppose  we  were  to  express  the  velocity  of  discharge  from  the  pipe 
line  in  Fig.  96  as  V  =  cv\/^2gh.     With  the  same  data  as  in  problem   (1) 
what  would  be  the  value  of  c*? 

Ans.  cv  =  0.1365. 

4.  What  will  be  the  pressure  at  a  point  1,000  ft.  from  the  end  of  the  pipe 
in  Fig.  96,  if  the  point  is  located  10  ft.  above  the  mouth  of  the  pipe?     The 
length  of  pipe  is  2,000  ft.,  the  diameter  is  6  in.,  and  the  value  of  h  is  50  ft. 

75.  Loss  of  Head  at  Entrance. — Whenever  the  velocity  of  a 
flowing  stream  is  abruptly  altered  there  will  be  eddy  currents  set 
up  which  will  cause  a  certain  loss  of  head  due  to  the  internal 
friction  of  the  particles  of  water  against  each  other.  Thus 
when  water  flows  into  a  pipe  from  a  reservoir,  the  loss  of 
head  within  the  first  few  feet  may  be  much  greater  than  that 
due  to  pipe  friction  alone  in  that  same  distance.  This  additional 


(a)    &= 


(c)   k'=i.Q 
FIG.  97.  —  Entrance  and  discharge  losses. 


(d)  k  =1.0 


loss  is  called  entrance  loss.  It  is  estimated  to  be  about  the  same 
as  the  loss  of  head  in  a  short  tube  such  as  those  shown  in  Fig.  71. 
Using  the  velocity  coefficients  which  are  there  given  and  de- 
termining values  of  k  from  them,  it  may  be  found  that  for  a  bell- 
mouthed  intake  the  entrance  loss  is  very  small  and  so  it  is 
usually  neglected.  For  a  pipe  that  is  flush  with  the  surface  of 
the  reservoir  wall,  such  as  Fig.  97(6),  the  value  of  k  is  taken  to 
be  0.5,  while  for  a  projecting  pipe,  such  as  in  Fig.  97 (c),  the  value 
of  k  is  assumed  to  be  1.0.  These  are  not  the  precise  values  that 
are  obtained  from  the  values  of  the  velocity  coefficients,  but  are 
close  enough  for  practical  purposes  when  it  is  realized  that  the 
entrance  losses  are  often  very  small  as  compared  with  the  other 
losses  of  head. 

If  V  indicates  the  velocity  in  the  pipe  itself,  the  losses  of  head 
at  entrance  may  be  assumed  to  be: 
For  a  non-projecting  pipe 

H'  =  0.5^  (61) 


100  HYDRAULICS 

For  a  projecting  pipe 

#'  =  1.0^  (62) 

76.  Loss  of  Head  at  Discharge. — In  the  case  of  a  pipe  dis- 
charging into  a  body  of  water  at  rest,  as  in  Fig.  97  (d),  the  entire 
kinetic  energy  of  the  stream  may  be  lost,1  for  considering  the 
body  of  water  in  Fig.  97 (d)  to  be  so  large  that  the  velocity  at  (2) 
is  negligible  we  may  write, 

#1  =  y  +  0  +  V  2/2g,        H2  =  y  +  0  +  0. 
Then  #'i_2  =  #1  -  H2 

or 

72 

H'--%  (63) 

t7 

77.  Loss  of  Head  in  Nozzle. — Although  a  nozzle  does  not 
produce  an  abrupt  change  of  velocity,  it  nevertheless  causes  a 


FIG.  98.  —  Loss  in  nozzle. 

certain  loss  of  head  by  virtue  of  which  its  velocity  coefficient  is 
less  than  unity.     For  Fig.  98  we  may  write  //'  =  H  i  —  HI.     Since 


l  ~  c,2  20 
and 


Therefore  for  the  nozzle 

H'  =  (^-,  -  l)  ~  (64) 

giving  for  k  the  value 

*='!-i 

cv 

exactly  as  in  the  case  of  the  orifice  in  Art.  52.  Note  that  in 
equation  (64)  the  loss  of  head  in  the  nozzle  is  based  upon  the 
jet  velocity.  % 

1  Unpublished  experimental  work  by  L.  F.  Moody  of  Rensselaer  Poly- 
technic Institute  indicates  that  in  some  cases  at  least  there  may  be  a  certain 
amount  of  diffusion  so  that  only  about  70  per  cent,  of  the  kinetic  energy  is 
lost,  the  rest  being  converted  into  pressure. 


FLOW  THROUGH  PIPES 


101 


78.  Other  Minor  Losses  of  Head. — When  there  is  an  abrupt 
contraction  of  the  stream  as  in  Fig.  99(a),  there  is  a  loss  of  head 
Hf  =  kVzz/2g,  where  k  has  the  values  given  in  the  following 
table:1 


TABLE  III 


F2 
Fi 

0.10 

0.20 

0.30 

0.40 

0.50 

0.60 

0.70 

0.80 

0.90 

k 

0.362 

0.338 

0.308 

0.267 

0.221 

0.164 

0.105 

0.053 

0.015 

When  there  is  an  abrupt  enlargement  of  the  stream  as  in 
Fig.  99(6),  theory  and  experiment  indicate  that  the  loss  of 
head  may  be  approximately  represented  by 

H'  =  (V,  -  V, 


Fi 


20 


(«)  (6) 

FIG.  99. — Losses  due  to  abrupt  change  in  cross-section. 


In  the  case  of  an  obstruction  in  a  pipe  the  loss  of  head  is 
usually  assumed  to  be  largely  that  due  to  expansion  after  passing 
through  the  constricted  part  and  hence  is  computed  by  the 
method  just  given. 

It  may  be  noted  that  all  of  the  losses  of  head  may  be  rep- 
resented as  some  function  of  the  velocity  head.  In  many  cases 
the  loss  may  be  small  as  compared  to  other  losses  in  the  same 
pipe  and  hence  its  omission  involves  little  error.  As  a  general 
rule  it  may  be  stated  that  any  feature  in  the  pipe  line  which 
disturbs  or  changes  the  velocity  of  flow  induces  some  additional 
friction  loss.  Exact  and  reliable  equations  or  coefficients  for 
many  of  these  losses  are  lacking  and  no  attempt  will  be  made  to 
give  more  of  them  here.2 

1L.  M.  Hoskins,  "Hydraulics,"  page  74.     After  data  from  Weisbach. 
2  See  Hughes  and  Safford,  "Hydraulics;"  Gibson,  "Hydraulics  and  Its 
Applications;"  Lea,  "  Hydraulics." 


102 


HYDRAULICS 


79.  Flow  through  Long  Pipe  Line.  —  The  strict  application  of 
the  general  equation 

HI  —  HZ  =  H 

would  require  us  to  express  Hr  as  a  function  of  all  the  various 
losses  that  might  exist  in  a  certain  pipe  line.  This  procedure 
is  followed  in  the  case  of  short  pipes  but  in  the  case  of  a  long 
pipe  line,  whose  length  is  at  least  one  thousand  diameters, 
it  will  usually  be  found  that  the  loss  in  pipe  friction  alone  renders 
the  others  insignificant.  For  we  have  just  seen  that  all  the 
losses  in  a  pipe  line  may  be  expressed  in  the  form 

V2 
H'  =  kTg 

and  in  the  case  of  entrance,  discharge,  and  other  similar  minor 
losses  values  of  k  are  either  less  than  unity  or  but  very  little 
greater.  But  for  pipe  friction  alone  we  have  seen  that 


and  if  I  is  only  great  enough  the  magnitude  of  the  quantity 
fl/d  may  make  all  other  values  of  k  negligible  by  comparison. 
In  view  of  the  uncertainty  of  the  exact  value  of  /  as  well  as 
other  loss  factors,  too  great  a  degree  of  refinement  is  unwarranted. 


FIG.  100. 

Assume  a  pipe  line  discharging  freely  into  the  air  at  (2)  in 
Fig.  100.     We  may  write  for  this 

Hl  =  h,  H,  =  FV2?. 

If  the  length  be  great  enough  all  other  losses  save  those  due  to 
pipe  friction  may  be  neglected  and  hence  between  (1)  and  (2), 

H'  =  f-j  TT-.     Thus  from  the  general  equation  we  have 
h~  2g  =  fd  Yg 


FLQW  THROUGH  PIPES  103 

/  l\V2 

Therefore   h  =  ( 1  +  /-r]  y.     However,  if  the  length  be  great 

enough  (  /  77 )  may  De  so  much  greater  than  unity  that  the  ex- 
pression 

h=fLdT9  <65> 

is  often  used.  But  it  should  be  noted  that  this  is  applicable  only 
when  the  pipe  line  is  long  and  when  the  velocity  of  the  stream  at 
(2)  is  no  greater  than  that  in  the  pipe  itself.  Equation  (65) 
is  equivalent  to  the  assumption  that  H'  =  h. 

Inspection  of  equation  (65)  shows  that  for  a  given  head  and 
length  of  pipe,  the  velocity  will  vary  somewhat  with  the  di- 
ameter of  the  pipe.  For  by  equation  (60)  /  decreases  as  d 
increases  and  in  (65)  the  ratio  l/d  also  becomes  smaller  with 
larger  diameters ;  therefore  the  entire  coefficient  of  V2/2g  becomes 
smaller  as  the  size  of  the  pipe  increases.  Hence  for  the  same 
value  of  h,  V  will  increase  as  the  diameter  of  the  pipe  increases, 
and  it  may  be  shown  that  V  varies  as  d0'5  to  °'6. 

EXAMPLES 

1.  Suppose  that  in  Fig.  100  the  pipe  projects  into  the  reservoir  at  entrance 
and  discharges  freely  into  the  air  at  (2),  the  size  of  the  jet  being  equal  to  the 
diameter  of  the  pipe.     If  h  =  40  ft.,  d"  =  12  in.,  and  I  =  50  ft.,  compute 
the  rate  of  discharge  considering  all  losses. 

Ans.  22.7  cu.  ft.  per  sec. 

2.  In  problem  (1)  if  I  =  1,000  ft.,  all  other  data  remaining  the  same, 
compute  the  rate  of  discharge  considering  all  losses.     Compute  the  rate  of 
discharge  by  the  approximate  method,  neglecting  minor  losses. 

Ans.  8.2  cu.  ft.  per  sec.;  8.56  cu.  ft.  per  sec. 

3.  Suppose  in  Fig.  100  that  a  nozzle  on  the  end  of  the  pipe  line  discharges 
a  jet  which  is  2.5  in.  in  diameter.     Assume  the  velocity  coefficient  of  the 
nozzle  to  be  0.95.     If  h  =  260  ft.,  d"  =  10  in.,  and  I  =  5,000  ft.,  find  the 
rate  of  discharge. 

Ans.  3.45  cu.  ft.  per  sec. 

4.  Find  the  rate  of  discharge  in  problem  (3)  if  the  nozzle  were  removed 
so  that  the  pipe  would  discharge  freely  a  stream  of  10  in.  diameter. 

Ans.  6.11  cu.  ft.  per  sec. 

6.  In  problem  (3)  find  the  pressure  head  in  the  pipe  at  the  base  of  the 
nozzle. 

Ans.  176.8ft. 

80.  Hydraulic  Gradient. — If  a  piezometer  tube  be  erected 
at  B  in  Fig.  101,  the  water  will  rise  in  it  to  some  height  BB' 
equal  to  the  pressure  head  existing  at  that  point.  If  the  lower 
end  of  the  pipe  were  closed  so  that  no  flow  could  occur,  the 


104 


HYDRAULICS 


height  of  this  column  would  evidently  be  BM .  The  drop  from 
M  to  Be  that  is  found  when  flow  takes  place  is  due  to  two  factors, 
one  of  these  being  that  the  water  at  B  has  acquired  a  velocity 
head  and  the  other  that  there  has  been  a  loss  of  head  through 
friction  between  the  reservoir  and  B. 

If  a  series  of  water  piezometers  were  erected  along  the  pipe 
line,  the  water  would  rise  in  them  to  various  levels.  The  line 
drawn  through  the  summits  of  such  an  imaginary  series  of 
water  columns  is  called  the  hydraulic  grade  line  or  the  hydraulic 
gradient.  It  is  seen  that  this  line  is  an  indication  of  the  pressure 


FIG.  101. — Hydraulic  gradient. 

variation  along  the  pipe.  Thus  at  any  point  the  vertical  dis- 
tance from  the  pipe  line  to  the  hydraulic  gradient  is  the  pressure 
at  that  point.  Since  at  C  this  distance  is  zero  it  follows  that  at 
C  the  pressure  is  atmospheric.  And  at  D  the  line  is  below 
the  pipe  indicating  that  at  the  point  in  question  the  pressure 
is  below  that  of  the  atmosphere  and  is  equal  to  —  DN.  The 
advantage  of  the  construction  of  the  hydraulic  gradient  is  that 
it  gives  a  very  clear  picture  of  the  pressure  variation  along  a 
pipe  line.  Also  in  practical  applications  the  profile  of  a  pro- 
posed pipe  line  should  be  drawn  to  scale.  Then  by  computing 
pressures  at  a  few  points  only  the  hydraulic  gradient  can  be 
drawn  and  from  it  the  pressures  at  all  points  can  be  quickly 
determined. 

The  hydraulic  gradient  is  not  necessarily  a  straight  line. 
For  a  pipe  of  uniform  diameter  it  will  be  a  straight  line  only  if- 
the  pipe  itself  is  straight,  but  for  ordinary  amounts  of  curvature 
the  hydraulic  gradient  will  deviate  very  slightly  from  a  straight 
line.  Of  course  if  there  are  losses  of  head  aside  from  those  due 
to  pipe  friction  there  will  be  abrupt  drops  in  the  hydraulic 
gradient,  and  any  variations  in  velocity  head  due  to  changes  in 
diameter  affect  the  hydraulic  gradient. 

It  may  be  seen  that  if  the  velocity  head  is  constant  the  drop 
in  the  hydraulic  gradient  between  any  two  points  is  the  measure 


FLOW  THROUGH  PIPES 


105 


of  the  loss  of  head  between  those  two  points.  But  if  the  velocity 
changes,  the  hydraulic  gradient  might  actually  rise  in  the  direc- 
tion of  flow. 

Additional  illustrations  of  the  hydraulic  gradient  are  given 
in  Figs.  102,  105,  108,  and  109. 


FIG.  102. — Hydraulic  gradient. 

EXAMPLES 

1.  Draw  the  hydraulic  gradient  for  problem  (3)  of  Art.  79. 

2.  What  would  the  hydraulic  gradient  be  for  problem  (4)  of  Art.  79? 

3.  In  Fig.  102  assume  the  pipe  to  be  of  a  uniform  diameter  of  20  in.  and 
500  ft.  in  length.     If  h  =  30  ft.,  find  the  distance  the  hydraulic  gradient 
drops  below  the  surface  of  the  water  for  a  point  just  within  the  pipe  at 
entrance.     Consider  all  losses.     Why  is  there  not  a  similar  drop  of  the 
hydraulic  gradient  at  the  point  of  discharge? 

Ans.  7.23  ft. 

4.  Suppose  there  were  no  discharge  loss  in  problem  (3),  what  would  have 
to  be  the  shape  of  the  hydraulic  gradient  near  the  discharge  end  of  the  pipe? 
What  would  have  to  be  the  nature  of  the  water  surface  above  the  mouth  of 
the  pipe  if  this  were  true? 

6.  A  pipe  line  4  in.  in  diameter  and  1,200  ft.  in  length  discharges  a  jet 
1  in.  in  diameter  at  a  point  300  ft.  below  the  surface  of  a  reservoir.  The 
velocity  coefficient  of  the  nozzle  is  0.90.  Find  the  pressure  head  in  the 
pipe  at  the  base  of  the  nozzle.  Draw  the  hydraulic  gradient. 

Ans.  232  ft. 

81.  Effect  of  Air  at  Summit. — In  Fig.  101  is  shown  a  pipe 
line  having  a  "summit"  at  D,  which  is  above  the  hydraulic 
gradient,  indicating  that  the  pressure  at  this  point  is  less  than 
atmospheric.  In  practice  this  would  be  avoided,  for  not  only 
might  the  excess  external  pressure  cause  this  portion  of  the 
pipe  to  collapse,  but  the  accumulation  of  air  at  this  point  might 
interfere  with  or  even  stop  the  flow  entirely.  All  ordinary  water 
carries  air  in  solution  and  readily  gives  it  up  at  a  point  of  low 
pressure  so  that  air  would  collect  in  time,  though  it  were  all 
expelled  by  some  means  in  the  beginning.  Therefore  in  design- 
ing a  pipe  line,  whenever  any  portion  of  it  is  found  to  be  above 


106 


HYDRAULICS 


the  hydraulic  gradient,  an  attempt  would  be  made  to  change 
the  profile  so  that  this  may  be  avoided.  In  case  this  is  impossible 
then  provision  must  be  made  for  exhausting  the  air  occasionally, 
if  full  flow  is  to  be  maintained. 

If  the  summit  is  below  the  hydraulic  gradient,  air  could  still 
collect,  though  not  so  readily  since  water  under  pressure  tends 
to  absorb  air.  But  under  such  conditions  it  is  very  easy  to 
release  the  air,  since  it  will  escape  if  an  opportunity  is  offered 
it.  A  valve  for  such  a  purpose  is  shown  in  Fig.  103.  Such 
valves  usually  have  a  float,  the  dropping  of  which,  as  air  collects 


Courtesy  Redwood  Manufacturers  Co. 

FIG.  103. — Air  valve  on  wooden  pipe  line. 

and  lowers  the  water  surface,  causes  a  valve  to  open.  When 
the  air  escapes,  the  water  level  rises  and  the  float  closes  the 
valve  again.  The  valve  in  Fig.  103  is  also  constructed  so  as  to 
admit  air  into  the  pipe  in  case  a  vacuum  should  accidentally 
occur  in  any  way.  This  will  prevent  the  pipe  from  collapsing 
in  such  an  event.  In  many  cases  it  is  highly  desirable  that  pipe 
lines  be  furnished  with  suitable  air  valves  for  both  these  purposes. 
In  Fig.  104  is  shown  how  a  vacuum  might  accidentally  occur, 
when  normally  the  pipe  is  under  a  positive  pressure.  We  have 
seen  that  the  greater  the  velocity  of  flow  through  a  pipe  line 
the  less  the  pressure  will  be  at  any  point.  Hence  if  some  event, 
such  as  the  bursting  of  the  pipe  at  C,  permits  a  larger  flow  of 
water,  the  hydraulic  gradient  will  be  much  steeper  than  normal. 
This  means  that  it  will  be  lowered,  and  it  may  be  lowered  suffi- 
ciently to  be  below  portions  of  the  pipe  as  in  Fig.  104. 


FLOW  THROUGH  PIPES  107 

Also  if  the  admission  of  water  to  a  pipe  line  is  shut  off  by 
the  closure  of  a  gate  valve  at  intake,  the  water  which  is  already 
in  the  pipe  will  tend  to  run  out.  As  no  more  water  can  get  in 
to  take  its  place,  a  vacuum  might  be  created  unless  air  were 
admitted.  Hence,  some  device  is  usually  provided  just  below 
the  valve  at  the  intake,  and  in  some  cases  at  other  points,  to 
admit  air  upon  such  an  occasion. 


82.  Hydraulic  Slope. — If  the  velocity  in  a  pipe  line  is  constant 
the  drop  in  the  hydraulic  gradient  is  equal  to  the  loss  of  head. 
The  ratio 

"'  (66) 


is  called  the  hydraulic  slope.  If  there  were  no  loss  of  head  the 
hydraulic  gradient  for  a  pipe  of  uniform  diameter  would  be  a 
horizontal  line.  Hence  the  steepness  of  this  line,  or  the  mag- 
nitude of  s,  is  a  measure  of  the  rate  of  loss. 

83.  Chezy's  Formula.  —  Equation  (58),  expressing  the  loss 
of  head  due  to  pipe  friction  may  often  be  written  and  used  in 
another  form  known  as  Chezy's  formula.  Thus  (58)  is 

tf'-/!!2 
~  4  m  2g 

Solving  this  equation  for  V2  we  have 


For  a  given  channel  Sg/f  may  be  a  constant  whose  value  we 
shall  indicate  by  c2  so  that 


(67) 

By  equation  (66)  H'/l  =  s,  the  slope  of  the  hydraulic  gradient, 
hence 

V  =  c  A/WS  (68) 


108  HYDRAULICS 

It  may  be  seen  that  equation  (68)  can  be  used  only  for  a  long 
pipe  line  to  which  equation  (65)  could  be  applied.  If  it  is 
desired  to  consider  other  losses  of  head  aside  from  pipe  friction 
Chezy's  formula  cannot  be  used. 

84.  Other  Formulas  for  Pipe  Friction. — It  has  already  been 
pointed  out  that  equation  (59)  is  known  not  to  be  correct  in 
form  but  is  widely  used  because  of  ease  of  computation,  its 
defects  being  covered  by  suitable  values  of  /.  The  true  equation 
is  of  the  form 

B'-f'^V-  (69) 

In  this  equation  /'  would  be  independent  of  both  d  and  V  and 
would  depend  only  upon  the  nature  of  the  pipe  wall.  The 
value  of  x  is  given  as  1.25,  though  occasionally  other  values 
differing  slightly  from  this  are  to  be  found.  The  value  of  n 
ranges  from  about  1.75  to  2.00  depending  upon  the  nature 
of  the  surface.  In  practical  cases  the  selection  of  /'  and  n  is 
about  as  difficult  as  the  choice  of /in  equation  (59).  On  account 
of  the  greater  ease  of  computation  with  equation  (59)  it  is  likely 
that  it  will  continue  to  be  used  and  that  more  precise  ways  of 
expressing/  will  be  determined.1 

In  regard  to  equation  (60)  it  has  been  established  that  values 
of  /  for  cast-iron  pipes  are  really  somewhat  less  than  given  by 
Darcy's  experiments,  but  this  discrepancy  should  be  looked 
upon  as  a  small  factor  of  safety.  The  formula  itself  errs  on  the 
side  of  safety. 

We  also  have  various  exponential  forms  of  Chezy's  formula, 
equation  (68),  and  in  these  also  the  exponents  are  variables 
but  the  coefficient  itself  depends  only  upon  the  nature  of  the 
surface.  The  one  most  commonly  used  in  this  country  is  the 
Hazen- Williams  formula  in  which  values  of  the  exponents  for 
average  conditions  are  used.  This  equation  is  of  the  form 

V  =  c'w°-63s°-54  (70) 

In   order  to  facilitate   computation   by  this  formula  a  set  of 
tables  has  been  prepared  and  also  a  special  slide  rule  constructed.2 

Moritz  gives  the  following  as  the  practice  of  the  U.  S.  Rec- 
lamation Service; 

q  =  C"d2-7(1000  s)°-555 

1  For  a  method  of  doing  this  see  article  by  R.  Biel,  Zeit.  des  Ver.  deut. 
Ing.,  June  27,  July  4,  1908. 

2  Williams  and  Hazen,  "Hydraulic  Tables." 


FLOW  THROUGH  PIPES  109 

where  c"  has  the  values  given  Table  IV.  Since  s  is  the  slope  of 
the  hydraulic  gradient,  1000  s  will  be  the  drop  of  the  hydraulic 
gradient  in  feet  per  thousand  feet  of  pipe. 

TABLE  IV 
Values  of  c" 


Wood  stave  pipe     

1.35 

Cast  iron  pipe               .        .... 

1.31 

Concrete  pipe 

1  24 

Riveted  steel  pipe  

1.18 

Although  the  Kutter  and  Bazin  formulas  for  determining  the 
coefficient  c  in  equation  (68)  were  intended  by  their  authors  to 
be  applied  to  open  channels,  they  have  been  widely  used  for 
closed  pipes  as  well.  This  has  been  due  to  the  fact  that  ex- 
perimental data  has  been  lacking  for  many  cases  of  pipe  lines, 
especially  in  large  sizes,  and  the  Kutter  and  Bazin  formulas  are 
of  a  general  nature  that  would  seem  to  make  them  fit  a  wide 
range  of  conditions.  These  formulas  are  given  in  the  following 
chapter. 

85.  Values  of  Friction  Factors. — Values  of  /  given  by  equation 
(60)  are  applicable  only  to  pipes  under  2  or  3  ft.  in  diameter 
since  the  largest  size  used  by  Darcy  in  his  experiments  was 
about  20  in.  It  has  also  been  stated  that  equation  (60)  gives 
values  of  /  that  are  now  known  to  be  somewhat  too  high  for 
smooth,  cast-iron  pipes,  but  are  conservative  values  that  can 
be  used  in  design.  In  Table  V  will  be  found  some  values  that 
are  more  nearly  correct,  though  these  values  are  for  large  pipes. 
The  form  of  equation  (60)  shows  that  for  large  pipes  the  value 
of  /  will  be  nearly  independent  of  the  diameter,  and  if  the  pipe 
surface  be  sufficiently  rough  for  the  loss  of  head  to  vary  as  V2, 
it  will  be  independent  of  the  velocity  also. 

Though  the  values  given  in  Table  V  are  for  large  pipes  only, 
they  will  serve  to  give  an  idea  of  the  correct  values  of  /  even 
in  the  case  of  small  pipes.  It  is  difficult  to  give  values  of  / 
that  can  be  applied  to  small  pipes  because  variations  in  the  rough- 
ness of  small  pipes  are  more  serious  in  their  effects.  Further- 
more the  walls  of  a  metal  pipe  become  covered  with  tubercles 
or  scaly  deposits  in  the  course  of  time.  After  a  depth  of  about  2 
in.  has  accumulated  this  action  ceases.  But  a  deposit  of  2  in. 
is  much  more  serious  in  reducing  the  capacity  of  a  small  pipe 
than  it  is  in  the  case  of  a  large  pipe. 


110 


HYDRAULICS 


Values  of  c  in  the  table  are  average  values  for  use  in  Chezy's 
formula,  equation  (68),  though  methods  of  computing  c  may  be 
found  in  the  next  chapter.  The  values  of  c'  in  Table  V  are 
for  use  in  the  Hazen- Williams  formula,  equation  (70).  These 
values  are  merely  selected  as  typical  for  the  classes  of  surface 
given  in  the  table.  The  values  of  n  in  Table  V  are  to  be  used 
in  the  formulas  of  Kutter  and  Manning,  which  are  to  be  found 
in  the  following  chapter. 

TABLE  V 


/ 

c 

c' 

n 

New  smooth   cast-iron  pipe         

0.015 

130 

170 

0.011 

^^ood-stave  pipe  new  or  old. 

0  018 

120 

160 

0.012 

New  riveted  pipe                                  

0.022 

110 

145 

0.013 

Old,  tuberculated,  cast-iron  pipe  or  riveted 
pine 

0.026 

100 

130 

0.014 

Any  old  and  rough  pipe 

0  040 

80 

105 

0.019 

86.  Size  of  Pipe  for  Given  Discharge. — It  is  possible  to  find 
the  diameter  of  pipe  necessary  for  a  given  rate  of  discharge  by 
direct  solution,  but  a  fifth  degree  equation  is  involved.1  It 
will  therefore  be  found  slightly  easier  and  simpler  to  solve  for 
the  diameter  by  a  method  of  "cut  and  try"  rather  than  to 
attempt  the  solution  of  the  higher  degree  equation,  though  the 
latter  is  not  difficult. 

The  procedure  of  "cut  and  try"  is  first  to  assume  any  diameter 
that  seems  reasonable  and  by  the  usual  methods  compute  the 
actual  rate  of  discharge  that  this  diameter  would  give.  Then 
compare  this  with  the  value  of  q  desired  and  if  it  is  too  large  or 
too  small  assume  a  new  size  of  pipe  and  repeat  until  the  computed 
rate  of  discharge  is  approximately  equal  to  that  specified.  Of 
course  one  should  use  only  commercial  sizes  of  pipe  and  not 
attempt  to  get  a  diameter  to  a  fraction  of  an  inch.  Naturally 
if  only  commercial  sizes  of  pipe  are  used  the  computed  rate  of 
discharge  may  not  agree  precisely  with  the  Value  desired,  and 
so  the  size  pipe  that  gives  the  nearest  to  that  should  be  chosen. 
If  the  conditions  of  the  problem  are  such  that  at  least  a  certain 
rate  of  discharge  must  bo  obtained,  then  a  size  pipe  should  be 


1  If  /  is  expressed  by  (60)  this  equation  will  be  of  the  sixth  degree. 


FLOW  THROUGH  PIPES  111 

used  which  will  give  a  slightly  greater  value  than  this  rather 
than  one  just  under  it.  In  the  problems  of  the  text  sizes  of 
pipe  are  used  in  whole  inches.  Actual  standard  pipe  dimensions 
are  given  on  page  263. 

If  good  judgment  is  used,  one  should  be  able  to  get  the  correct 
answer  within  two  or  three  trials  at  most.  In  order  to  do  this 
it  is  necessary  to  carefully  compare  the  rate  of  discharge  com- 
puted with  the  value  specified  and  then  to  estimate  how  much 
the  area  of  the  pipe  might  need  to  be  increased  or  diminished 
to  yield  the  proper  result.  In  doing  this  it  must  be  borne  in 
mind  that  the  velocity  is  not  the  same  in  pipes  of  different 
sizes.  The  larger  the  pipe  the  higher  the  velocity  of  flow  and 
hence  it  will  discharge  more  in  proportion  to  its  area  than  a 
smaller  pipe.  This  is  not  a  matter  that  is  worth  making  any 
computations  on  for  this  purpose,  but  it  might  be  noted  that, 
all  other  things  being  equal,  the  rate  of  discharge  varies  about 
as  d2-6.  Therefore  in  assuming  a  new  diameter  one  should 
not  go  quite  so  far  from  the  former  value  as  one  would  if  the 
velocity  remained  the  same  in  value,  in  which  case  q  would 
vary  as  d2. 

As  an  example,  suppose  it  is  desired  to  find  the  diameter  of 
pipe  necessary  to  discharge  9  sec.  ft.  with  a  total  fall  of  50  ft. 
in  a  distance  of  2  miles.  Since  the  length  is  so  great  we  may  use 
the  approximate  formula,  equation  (65).  Suppose  we  assume 
d"  =  10  in.,  then  /  =  0.0220  and 

V2       50 

~ 


10          2g 
Solving  this  we  find  that 

V  =  3.40  ft.   per  sec. 
and 

q  =  0.545  X  3.40  =  1.85  sec.  ft. 

The  value  desired,  of  9  sec.  ft.,  is  9/1.85  or  about  5  times  the 
result  obtained.  If  the  velocity  were  unchanged  we  should 
require  a  pipe  whose  area  was  5  times  that  of  a  10-in.  pipe.  But 
we  know  that  the  velocity  will  be  somewhat  larger  with  a  larger 
pipe,  and  so  we  would  assume  a  pipe  whose  area  is  about  4  times 
that  of  the  10-in.  pipe,  trusting  that  the  increase  in  velocity  will 
make  up  the  difference.  Now  this  pipe  would  have  a  diameter 
the  square  of  which  would  be  about  4  times  102  .  or  400. 


112  HYDRAULICS 

So  we  use  20  in.  as  a  second  trial  value.     For  a  20-in.  pipe, 
/  =  0.0210  and 


. 

Solving  this  we  obtain 

V  =  4.92  ft.   per  sec. 
and 

q  =  2.18  X  4.92  =  10.74  sec.  ft. 

This  answer  would  appear  to  be  satisfactory  as  it  is  on  the  safe 
side  and  is  not  much  larger  than  the  value  desired.  The  next 
size  pipe  below  this  is  18  in.  and  we  know  that  the  rate  of  dis- 

182 
charge  through  it  must  be  less  than  ^  X  10.74  =  8.7.     (The 

actual  value  for  the  18-in.  pipe  is  8.20.)  Hence  we  would 
conclude  that  a  20-in.  pipe  would  be  required  unless  it  were 
allowable  for  the  capacity  to  be  somewhat  less  than  9  sec.  ft. 

EXAMPLE 

1.  What  must  be  the  diameter  of  pipe  necessary  to  discharge  6.5  sec. 
ft.  under  a  head  of  120  ft.  if  the  length  of  pipe  is  5,000  ft. 

87.  Power  Delivered  by  a  Pipe.  —  In  Fig.  105  consider  a 
point  C  which  is  located  near  the  end  of  a  pipe  line.  When 
no  flow  occurs,  due  to  the  closure  of  a  valve  or  other  device 


c 
FIG.   105. — Varying  hydraulic  gradient  with  different  rates  of  discharge. 

beyond  C,  the  pressure  at  C  is  a  maximum,  being  equal  to  CX. 
But  when  flow  occurs  the  pressure  at  C  drops  to  the  value  CY, 
and  the  greater  the  rate  of  discharge  the  steeper  will  be  the 
hydraulic  gradient  and  the  less  will  be  the  pressure  at  C.  If 
the  nozzle,  or  other  device  beyond  C,  be  removed  entirely  making 
C  a  point  at  the  very  end  of  the  pipe,  the  pressure  will  then 
be  reduced  to  zero.  In  Fig.  106  are  shown  the  decrease  in 
pressure  head  at  C  and  the  increase  in  velocity  head  at  C  as 


FLOW  THROUGH  PIPES 


113 


the  rate  of  discharge  is  caused  to  increase  by  opening  wider 
whatever  device  is  below  C.  Now  the  total  head  at  C  is  the 
sum  of  the  pressure  head  and  the  velocity  head,  but  it  is  seen 
to  continually  decrease  with  increasing  discharge  until  it  reaches 
a  minimum  value  which  is  the  velocity  head  when  the  pipe  is 
wide  open. 

We  have  seen  that  power  is  a  function  of  both  q  and  H  and 
may  be  expressed  as  wqHc-  In  the  case  under  consideration 
H  decreases  as  q  increases.  When  q  is  zero  Hc  is  a  maximum 


Rate  of  Discharge 

FIG.  106. — Head  and  power  delivered  by  a  pipe. 

but  the  power  is  zero.  And  when  q  is  a  maximum  the  power  is 
small  due  to  the  small  value  of  HC-  Somewhere  between  these 
two  extremes  the  product  of  these  two  variables  reaches  a 
maximum  as  shown  by  Fig.  106.  It  can  be  shown  that  the  power 
is  a  maximum  when  the  flow  is  such  that  one-tKird  of  the  static 
head  is  consumed  in  friction. 

The  efficiency  of  a  pipe  line  may  be  defined  as  the  ratio  of  the 
power  delivered  to  the  power  supplied.  But  power  is  propor- 
tional to  head,  and  hence  the  efficiency  is  Hc/h,  where  h  =  CX  in 
Fig.  105.  In  the  case  of  maximum  power  delivered  one-third  of 
CX  has  been  consumed  in  friction,  hence  the  efficiency  is  only 
66%  per  cent.  If  economy  of  water  is  no  object  it  would  be 
desirable  to  transmit  power  under  these  conditions  as  the  cost 
of  the  pipe  line  would  be  small  in  proportion  to  the  power  de- 
livered. But  under  usual  conditions  it  is  undesirable  that 
one-third  of  the  energy  of  the  water  be  wasted,  and  hence  such  a 
size  of  pipe  line  would  be  employed  that  it  could  deliver  the 
water  available  with  a  loss  of  only  a  few  per  cent.  With  ordinary 


114 


HYDRAULICS 


power-plant  practice  the  efficiency  of  the  pipe  lines  leading  to 
the  turbines  is  about  95  per  cent. 

EXAMPLES 

1.  Find  the  power  delivered  in  the  jet  in  problem  (3)  of  Art.  79. 

2.  What  is  the  efficiency  of  the  pipe  line  (and  nozzle)  in  problem  (3)  of 
Art.  79? 

Ans.  61.5  per  cent. 

3.  A  pipe  line  2,000  ft.  long  is  5  ft.  in  diameter.     If  the  fall  from  the 
reservoir  to  the  end  of  the  pipe  is  120  ft.,  what  is  the  maximum  amount  of 
power  the  pipe  could  deliver? 

4.  What  amount  of  power  would  the  pipe  in  problem  (3)  deliver  if  its 
efficiency  were  95  per  cent.? 

.     6.  What  size  pipe  would  be  required  to  deliver  the  water  discharged  in 
problem  (3)  if  the  efficiency  of  the  pipe  were  to  be  90  per  cent.? 

88.  Pipe  Line  with  Pump. — In  case  a  pump  lifts  water  from 
one  reservoir  to  another,  as  in  Fig.  107,  it  not  only  does  work  in 
lifting  the  water  the  height  z  but  it  also  has  to  overcome  the 


FIG.   107. — Pipe  line  with  pump. 


friction  loss  in  the  suction  and  discharge  piping.  This  friction 
head  is  equivalent  to  some  added  lift  so  that  the  effect  is  the 
same  as  if  the  pump  lifted  the  water  a  height  z  +  Hf,  without 
loss.  Hence  the  power  delivered  to  the  water  by  the  pump  is 

W(z  +  H')  (71) 

The  power  required  to  run  the  pump  is  greater  than  this,  de- 
pending upon  the  efficiency  of  the  pump.  Although  the  pump 
actually  lifts  the  water  a  height  z,  it  is  said  to  work  against  a 
head  h  whose  value  is 

h  =  z  +  H'  (72) 


FLOW  THROUGH  PIPES  115 

In  case  the  pump  discharges  a  stream  of  water  through  a 
nozzle,  such  as  in  Fig.  108,  the  water  has  not  only  been  lifted 
a  height  z  but  it  has  also  received  kinetic  energy  proportional 
to  Vzz/2g,  where  Vz  is  the  velocity  of  the  jet.  Thus  the  power 
delivered  to  the  water  by  the  pump  is 

w(,  +  ^  +  H')  (73) 

And  the  head  against  which  the  pump  works  is  now 

*-.  +  £  +  *'  (74) 

The  difference  between  the  two  cases  in  Figs.  107  and  108  is 
really  slight.     In  equation  (71)  we  have  considered  the  velocity 


FIG.  108. — Pipe  line  with  pump. 

head  at  E  to  have  been  lost,  while  in  equation  (73)  the  velocity 
head  in  the  jet  has  not  yet  been  lost.  Thus  Hf  in  (71)  includes 
the  velocity  head  of  discharge  while  the  H'  in  (73)  does  not. 

EXAMPLES 

1.  A  10-in.  pipe  line  is  3  miles  long.     If  4  cu.  ft.  of  water  per  sec.  are  to  be 
pumped  through  it,  the  total  actual  lift  being  20  ft.,  what  will  be  the  horse- 
power required  if  the  pump  efficiency  is  70  per  cent.  ? 

Ans.  240  hp. 

2.  In  Fig.  107  assume  d"  =  10  in.,  BC  =  20  ft.,  DE  =  3,000  ft.,  and 
z  =  135  ft.     If  q  =  7  sec.  ft.  and  the  pump  efficiency  is  80  per  cent.,  what 
is  the  power  required? 

Ans.  340  hp. 

3.  In  problem  (2),  if  the  elevation  of  C  above  the  water  surface  is  13  ft. 
and  that  of  D  is  15  ft.,  compute  the  pressures  at  C  and  D. 

Ans.  pc  =  -19.48  ft.,  pD  =  +323  ft. 

4.  In  Fig.  107  assume  d"  =  3  in.,  BC  =  20  ft.,  DE  =  200  ft.,  and  z  =  70 
ft.     The  elevation  of  C  above  the  water  surface  is  15  ft.     (a)  If  the  pressure 
at  C  is  to  be  —25  ft.,  what  is  the  rate  at  which  water  is  pumped?     (6)  If 
the  efficiency  of  the  pump  is  60  per  cent.,  what  is  the  power  required? 

Ans.  (a)  q  =  0.613  cu.  ft.  per  sec.     (6)  15  hp. 


116 


HYDRAULICS 


5.  When  a  certain  pump  is  delivering  1.0  cu.  ft.  of  water  per  sec.,  the 
pressure  gage  at  D  (Fig.  107)  reads  20  Ib.  per  sq.  in.,  while  a  vacuum  gage 
at  C  reads  10  in.  of  mercury.  The  pressure  gage  is  2  ft.  higher  than  the 
vacuum  gage.  If  the  diameter  of  the  suction  pipe  is  4  in.  and  that  of  the 
discharge  pipe  is  3  in.,  find  the  power  delivered  to  the  water. 

Ans.  7.23  hp. 

89.  Pipe  Line  with  Turbine. — The  type  of  machine  that 
is  usually  employed  for  converting  the  energy  of  water  into 
mechanical  work  is  called  a  turbine.  In  flowing  from  the 
upper  body  of  water  in  Fig.  109  to  the  lower,  the  water  loses  its 
potential  energy  due  to  the  elevation  z.  This  energy,  which  the 


FIG.  109. — Pipe  line  with  turbine. 

water  loses,  is  expended  in  two  ways.  A  part  of  it  is  consumed  in 
hydraulic  friction  in  the  pipe  and  the  rest  of  it  is  delivered  to 
the  turbine.  Of  that  which  is  delivered  to  the  turbine,  a  portion 
is  lost  in  hydraulic  friction  within  the  machine  and  the  rest  is 
converted  into  mechanical  work. 

The  power  delivered  to  the  turbine  is  decreased  by  the  fric- 
tion loss  in  the  pipe  line,  and  its  value  is  given  by 

W(z  -  H')  (75) 

The  power  delivered  by  the  machine  is  less  than  this  depending 
upon  both  the  hydraulic  and  mechanical  losses  of  the  turbine. 
The  head  under  which  the  turbine  operates  is 


z  -  H1 


(76) 


In  the  case  of  a  turbine  the  only  loss  of  head  Hf,  which  is  de- 
ducted, is  that  in  the  supply  pipe.  The  draft  tube,  as  the 
conduit  which  leads  the  water  away  from  the  turbine  is  called, 
is  considered  an  integral  part  of  the  machine  and  hence  h  should 
cover  losses  in  it  as  well  as  in  the  turbine  case  itself. 


FLOW  THROUGH  PIPES  117 

In  applying  these  equations  it  should  be  noted  that  the 
particular  location  of  the  turbine  is  immaterial  so  long  as  it 
is  not  set  so  high  above  the  lower  water  level  that  the  pressure 
at  the  top  of  the  draft  tube  approaches  absolute  zero  in  value. 
But  as  long  as  this  is  avoided  the  turbine  can  make  use  of  the 
entire  fall  to  the  lower  water  level  by  the  use  of  an  air-tight 
draft  tube.  The  higher  the  turbine  is  situated,  within  the  limit 
specified,  the  less  the  pressure  will  be  at  intake  but  this  is  offset 
by  an  increased  suction  on  the  discharge  side. 

EXAMPLES 

1.  In  Fig.  109  assume  d"  =  12  in.,  EC  =  200  ft.,  and  z  =  120  ft.     The 
entrance  to  the  pipe  at  the  intake  is  flush  with  the  wall,     (a)  If  q  =  8  sec. 
ft.,  what  is  the  head  supplied  to  the  turbine?     (6)  What  is  the  power  deliv- 
ered by  the  turbine  if  its  efficiency  is  75  per  cent.? 

Ans.   (a)   h  =  112.2  ft.     (6)  76.5  hp. 

2.  In  problem  (1)  if  the  elevation  of  C  above  the  water  level  is  18  ft., 
what  is  the  pressure  head  at  that  point? 

Ans.   92.56  ft. 

3.  A  turbine  operating  under  a  total  fall  of  120  ft.  (z  =  120  ft.),  is  sup- 
plied with  water  through  300  ft.  of  8-in.  pipe.     If  the  rate  of  discharge  be 
such  that  30  ft.  of  head  is  lost  in  friction  in  the  pipe,  what  will  be  the  power 
delivered  to  the  turbine? 


From  a  photograph  by  the  author. 

FIG.  110. — Cast-iron  pipe  line. 

90.  Construction  of  Pipe  Lines. — Cast-iron  pipes  have  been 
employed  for  the  last  200  years  and  are  very  satisfactory  for 
ordinary  water-works  purposes  where  moderate  heads  are 
employed.  They  are  very  durable  and  require  but  little  atten- 


118 


HYDRAULICS 


tion.  While  it  is  sometimes  used  under  hi'gher  pressures,  cast 
iron  is  not  considered  desirable  for  heads  above  400  ft.  nor  is  it 
suitable  for  very  large  diameters  on  account  of  low  tensile 
strength  and  possible  defects  in  casting.  For  temporary  pur- 


From  a  photograph  by  the  author. 

FIG.  111. — Riveted  steel  pipe  under  head  of  1300  ft.  leading  to  Drum  power 
house  of  Pacific  Gas  &  Elec.  Co.  in  California. 

poses  or  for  cheaper  installations  pipes  are  sometimes  made 
of  very  light  weight  riveted  steel,  usually  coated  with  some 
material  in  order  to  enable  them  to  resist  corrosion. 


From  a  photograph  by  the  author. 

FIG.  112. — Old  wooden  water  pipe  at  New  Orleans  made  from  cypress  log. 

For  high  pressures  cast  iron  is  unsuitable  and  steel  pipe  is 
used.  These  may  be  riveted  as  in  Fig.  Ill,  or  they  may  be 
welded  in  special  cases.  Riveted  steel  pipe  offers  more  resis- 
tance to  flow  than  a  new  cast-iron  pipe  on  account  of  the 


FLOW  THROUGH  PIPES  119 

projecting  rivet  heads  and  the  overlapping  of  the  plates,  but 
an  old  riveted  steel  pipe  and  an  old  cast-iron  pipe  are  about  the 
same  since  both  become  coated  alike  with  tubercles.  A  steel 
pipe  is  not  considered  as  durable  as  a  cast-iron  pipe,  but  for 
high  heads  it  is  necessary  to  use  it. 

For  heads  under  200  or  300  ft.  wood-stave  pipe  offers  many 
advantages.  It  is  cheaper  than  a  metal  pipe  for  the  same 
service.  The  resistance  to  flow  is  less  than  a  riveted  steel  pipe 


Courtesy  of  Redwood  Manufacturers  Co. 

FIG.   113. — Construction  of  wood-stave  pipe. 

and  about  the  same  as  a  new,  smooth,  cast-iron  pipe,  but  it  has 
the  advantage  that  its  capacity  does  not  decrease  with  age. 
The  early  types  of  wooden  pipe  used  were  simply  hollow  logs 
as  shown  in  Fig.  112.  Some  of  these  were  used  for  many  years. 
Modern  wood  pipe  is  generally  built  up  of  staves  as  shown  in 
Fig.  113.  The  staves  are  so  arranged  that  the  joints  are 
"broken."  In  order  to  make  a  water-tight  joint,  a  thin  steel 
tongue  is  inserted  in  a  saw  cut  across  the  end  of  each  stave. 
This  piece  of  steel  is  slightly  wider  than  the  stave  so  that  when 
the  bands  are  tightened  up  it  will  sink  into  the  staves  on  either 
side  a  distance  of  about  in.  or  more.  In  the  life  of  a  wood- 


120 


HYDRAULICS 


stave  pipe  the  encircling  metal  bands  often  have  to  be  renewed. 
It  is  essential  that  a  wooden  pipe  be  kept  filled  with  water,  if 
it  is  to  have  a  long  life,  as  wood  does  not  rot  rapidly  if  it  is  kept 
continually  wet  or  continually  dry.  It  rots  the  worst  when  it 
is  exposed  to  alternations  in  these  conditions.  The  life  of  a 
wood  pipe  is  not  as  long  as  that  of  a  heavy  cast-iron  pipe  but  it 


From  a  photograph  by  the  author. 

FIG.  114. — Curves  in    wood-stave  pipe.     In  the  Sierra  Nevada   Mts.    of 

California. 

may  be  as  long  as  that  of  a  steel  pipe.  However,  statistics  of 
these  matters  are  lacking  and  subject  to  much  dispute.  The 
wood  pipe  is  free  from  corrosion  and  from  electrolysis  and  is  not 
attacked  by  acids  in  the  water.  Hence,  it  is  often  used  for 
carrying  liquids  that  could  not  be  handled  by  a  metal  pipe. 
It  is  possible  to  introduce  broad  sweeping  curves  into  a  wood- 


FLOW  THROUGH  PIPES  121 

stave  pipe  without  any  special  devices  or  fittings,  as  shown  in 
Fig.  114. 

Metal  pipes  are  subject  to  expansion  and  contraction  due  to 
temperature  changes  and  provision  must  often  be  made  for  this. 
In  the  case  of  a  cast-iron  pipe  line  the  amount  of  play  afforded 
at  each  joint  is  usually  sufficient.  But  a  riveted-steel  pipe  line 
has  no  such  flexibility  and  expansion  devices  may  be  employed. 


From  a  photograph  by  the  author. 
FIG.  115. — Expansion  joint  in  8.5  ft.  riveted  steel  pipe  under  low  head. 

One  type  of  expansion  joint  is  shown  in  Fig.  115,  which  is  suit- 
able only  for  low  pressures.  It  may  be  seen  that  the  circular 
plates  can  spring  enough  to  permit  the  necessary  endwise  motion 
of  the  pipe.  For  higher  pressures  a  joint  such  as  in  Figs.  116 
and  117  may  be  used. 

When  water  is  lifted  by  a  pipe  line  to  a  greater  height  than 
the  initial  water  level,  as  in  Fig.  118,  the  pipe  is  called  a 
siphon.  Of  course  it  is  necessary  to  exhaust  the  air  by  some 
means  in  order  to  start  the  flow,  and  if  the  flow  is  to  continue 


122 


HYDRAULICS 


From  a  photograph  by  the  author. 
FIG.   116. — Expansion  joint  in  high  pressure  riveted  steel  pipe  line. 


FIG.  117. — Expansion  joint. 


FLOW  THROUGH  PIPES 


123 


the  air  which  collects  at  the  summit  must  be  removed  from  time 

to  time.     There  are  times  when  such  a  device  cannot  be  avoided. 

By  analogy  a  pipe  line  such  as  shown  in  Fig.  119  is  called  an 

"  in  verted  siphon,"  and  it  is  usually  found  where  it  is  necessary 


FIG.  118.— Siphon. 


FIG.  119. — Inverted  siphon. 


Jin  a  photograph  by  the  author. 
FIG.  120. — Riveted  steel  siphon.     Lake  Spaulding  development  of  Pacific 

Gas  &  Elec.  Co. 

to  carry  water  across  a  valley  or  depression  as  in  Fig.  120. 
However,  it  is  quite  common  to  call  a  pipe  so  situated  simply 
a  "siphon." 


124  HYDRAULICS 

91.  PROBLEMS 

1.  A  pipe  line  850  ft.  long  discharges  freely  into  the  air  under  a  fall  of 
40  ft.     (Assume  a  projecting  pipe  at  entrance.)     (a)  If  d"  =    6  in.,  find  the 
rate  of  discharge.     (6)  If  d"  =  12  in.,  find  the  rate  of  discharge. 

Ans.   (a)  q  =  1.55  sec.  ft.     (b)  q  =  8.85  sec.  ft. 

2.  Suppose  a  pipe  line  runs  from  one  reservoir  to  another,  both  ends  of 
the  pipe  being  under  water.     Assume  the  intake  end  is  non-projecting.     If 
the  difference  in  water  levels  is  110  ft.,  the  length  of  pipe  is  500  ft.,  and  the 
diameter  10  in.,  what  will  be  the  rate  of  discharge?     What  will  the  capacity 
be  when  the  pipe  is  old? 

3.  In  problem  (2)  a  point  C  is  located  120  ft.  below  the  surface  of  the 
water  in  the  upper  reservoir  and  300  ft.  from  the  intake  end  of  the  pipe. 
What  will  be  the  pressure  at  that  point?     Draw  the  hydraulic  gradient. 

Ans.  49.5ft. 

4.  A  pipe  line  800  ft.  long  discharges  freely  at  a  point  150  ft.  below  the 
water  level  at  intake.     The  pipe  projects  into  the  reservoir.     The  first 
500  ft.  is  12  in.  in  diameter  and  the  remaining  300  ft.  is  8  in.  in  diameter. 
Find  the  rate  of  discharge. 

Ans.  q  =  9.25  sec.  ft. 

5.  The  junction  of  the  two  sizes  of  pipe  in  problem  (4)  is  120  ft.  below  the 
surface  of  the  water  level.     Find  the  pressure  just  above  C  and  just  below 
C,  where  C  denotes  the  point  of  junction.     Assume  a  sudden  contraction 
at  this  point. 

6.  A  jet  of  water  is  discharged  through  a  nozzle  at  a  point  200  ft.  below 
the  water  level  at  intake.     The  jet  is  4  in.  in  diameter  and  the  velocity 
coefficient  of  the  nozzle  is  0.90.     If  the  pipe  line  is  12  in.  in  diameter,  500  ft. 
long,  with  a  non-projecting  entrance,  what  is  the  pressure  at  the  base  of 
the  nozzle? 

Ans.   177.8ft. 

7.  It  is  desired  to  deliver  3  cu.  ft.  of  water  per  sec.  at  a  point  10,000  ft. 
distant  with  a  loss  of  head  of  150  ft.     What  size  pipe  would  be  required? 

8.  What  would  be  the  probable  capacity  of  the  pipe  in  problem  (7)  when 
it  was  old? 

9.  A  pump  delivers  water  through  300  ft.  of  4-in.  fire  hose  to  a  nozzle 
which  throws  a  1-in.  jet.     The  velocity  coefficient  of  the  nozzle  is  0.98  and 
the  value  of  /  for  the  hose  may  be  assumed  to  be  0.025.     The  nozzle  is  20 
ft.  higher  than  the  pump.     It  is  required  that  the  velocity  of  the  jet  be  70  ft. 
per  sec.     What  will  be  the  necessary  pressure  at  the  pump? 

10.  The  steel  siphon  shown  in  Fig.  120  is  8.5  ft.  in  diameter.     It  is  1,900 
ft.  long  and  carries  300  cu.  ft.  of  water  per  sec.     What  must  be  the  difference 
in  water  level  at  the  two  ends?      (It  is  arranged  as  in  Fig.  119.) 

11.  The  pipe  line  shown  in  Figs.  116  and  191  has  an  average  diameter 
of  62  in.,  is  6,272  ft.  long,  and  the  difference  in  level  between  the  power 
house  and  the  intake  is  1,375  ft.     When  the  pipe  delivers  300  cu.  ft.  of 
water  per  sec.,  what  is  its  efficiency? 

12.  What  is  the  horsepower  delivered  to  the  plant  in  problem  (11)? 


CHAPTER  VIII 
UNIFORM  FLOW  IN  OPEN  CHANNELS 

92.  Open  Channels. — An  open  channel  is  one  in  which  the 
stream  is  not  completely  enclosed  by  solid  boundaries  and  there- 
fore has  a  free  surface  subjected  only  to  atmospheric  pressure. 
The  flow  in  such  a  channel  is  not  dependent  upon  some  external 
head  but  rather  upon  the  slope  of  the  channel  and  of  the  water 
surface. 

The  principal  types  of  open  channels  are:  natural  streams  or 
rivers,  artificial  canals,  and  sewers,  tunnels,  or  pipe  lines  not 
completely  filled. 


From  a  photograph  by  the  author. 

FIG.  121. — Canal  of  the  Pac.  Gas  &  Elec.  Co.  with  one  bank 
rock  lined. 

The  accurate  solution  of  problems  of  flow  in  open  channels  is 
much  more  difficult  than  in  the  case  of  pressure  pipes.  Not  only 
is  reliable  experimental  data  more  difficult  to  secure,  but  there  is 
a  wider  range  of  conditions  than  is  met  with  in  the  case  of  pipes. 
Practically  all  pipes  are  round,  but  the  cross-sections  of  open 
channels  may  be  of  any  shapes  from  circular  to  the  irregular 
forms  of  natural  streams.  It  is  probable  that  the  shape  of  the 

125 


126 


HYDRAULICS 


cross-section  affects  the  flow  in  a  way  that  is  not  covered  by  the 
factor,  w,  the  hydraulic  mean  depth  (see  page  97).  In  pipes  the 
degree  of  roughness  ordinarily  ranges  from  that  of  new,  smooth, 


FIG.  122. 


cast-iron  or  wood-stave  pipes  on  the  one  hand  to  that  of  old 
corroded  pipes  on  the  other.  But  with  open  channels  the  sur- 
faces vary  from  smooth  timber  (Fig.  123)  to  the  rough  and 


From  a  photograph  by  the  author. 

FIG.  123. — Non-uniform  flow  in  wooden  flume. 

irregular  beds  of  some  rivers.  Hence  the  choice  of  friction  fac- 
tors is  attended  with  greater  uncertainty  in  the  case  of  open 
channels  than  in  the  case  of  pipes. 


UNIFORM  FLOW  IN  OPEN  CHANNELS 


127 


93.  Uniform  Flow. — If  the  shape  and  size  of  any  water  cross- 
section  is  identical  with  that  of  every  other  section  in  the  length 
of  channel  under  consideration,  the  flow  is  said  to  be  uniform. 
Such  cases  are  shown  in  Figs.  54  and  121.  Uniform  flow  must 
not  be  confused  with  steady  flow.  The  former  requires  that  the 
conditions  at  any  time  be  the  same  from  place  to  place;  the 


From  a  photograph  by  the  author. 

FIG.   124. — Cascade  on  Los  Angeles  aqueduct. 

latter  requires  that  the  conditions  at  every  section  be  constant 
with  respect  to  time.  We  might  have  steady  flow  for  both  uni- 
form and  non-uniform  flow  as  shown  in  Fig.  122.  Uniform  flow 
is  obtained  only  when  a  channel  is  uniform  for  a  considerable 
distance  so  that  the  water  has  a  chance  to  adjust  itself.  The 
channel  in  Fig.  123  is  uniform  but  the  flow  is  non-uniform  in  the 
portion  shown  because  the  water  has  just  entered  it  and  has  not 


128  HYDRAULICS 

yet  attained  a  condition  of  equilibrium.  The  conditions  are 
similar  to  the  upper  portion  of  the  channel  shown  in  Fig.  122.  On 
the  other  hand  the  flow  is  non-uniform  in  Fig.  124  because  the 
slope  of  the  channel  varies. 

94.  Hydraulic  Gradient. — It  is  quite  evident  that  in  the  case  of 
an  open  channel  the  hydraulic  gradient  coincides  with  the  water 
surface.     For  if  a  piezometer  tube  be  attached  to  the  side  of  the 
channel  the  water  will  rise  in  it  until  its  surface  is  level  with  that 
of  the  water  in  the  channel. 

95.  Equation  for  Uniform  Flow. — The  equation  that  is  most 
generally  used  for  steady  uniform  flow  in  open  channels  is  one 
that  is  also  often  used  for  flow  in  long  pipes.     It  is 

V  =  c  \^rns  (68) 

the  derivation  of  which  was  given  in  Art.  83.  In  this  formula,  c 
is  a  coefficient  dependent  upon'the  roughness  of  the  surface  in 
contact  with  the  water,  and  it  is  also  often  given  as  a  function  of 
other  variables  as  well.  The  quantity,  m,  is  the  hydraulic  mean 
depth,  or  hydraulic  radius,  and  its  value  is  given  by 

area  of  water  cross-section 


length  of  wetted  perimeter1 

In  the  case  of  an  open  channel,  s  is  the  slope  of  the  water  surface, 
according  to  Art.  94. 

Recognizing  that  the  velocity  does  not  vary  exactly  as  the 
square  root  of  m  or  of  s,  exponential  formulas  such  as  equation 
(70)  are  sometimes  used.  But  if  equation  (68)  is  employed  it  is 
seen  that  c  must  then  be  a  function  of  m  and  s,  since  equation 
(68)  does  not  involve  the  correct  exponents  of  m  and  s. 

96.  Kutter's  Formula  for  c. — The  formula  for  c  that  has  prob- 
ably been  more  widely  used  than  any  other  is  that  of  Kutter  and 
Ganguillet,  two  Swiss  engineers.  This  formula  is  based  upon  a 
wealth  of  data  from  small  artificial  canals  up  to  natural  streams 
as  large  as  the  Mississippi,  and  for  this  reason  it  is  believed  to  be 
applicable  to  a  wide  range  of  conditions.  But  any  formula  that 
attempts  to  cover  too  large  a  field  must  necessarily  be  a  mere 
average  of  a  number  of  scattered  values  and,  though  giving 
approximate  values  at  least  for  any  combination  of  factors,  it 

1  The  wetted  perimeter  is  only  that  portion  of  the  channel  section  that  is  in 
contact  with  the  water.  The  width  across  the  free  surface  of  the  water 
should  not  be  included. 


UNIFORM  FLOW  IN  OPEN  CHANNELS 


129 


cannot  be  expected  to  give  exact  values  in  individual  cases. 
Hence  too  great  reliance  must  not  be  placed  upon  values  given 
by  the  use  of  this  or  any  other  such  empirical  formula. 

In  Art.  95,  it  was  pointed  out  that,  since  equation  (68)  is  not  a 
true  expression  of  the  law  of  flow,  the  value  of  c  must  be  a  func- 
tion of  m  and  s  as  well  as  the  roughness  of  the  surface.  The 
formula  of  Kutter  takes  all  three  factors  into  consideration.  It 
is 


41.65  + 


+ 


(41.65  + 

V 


(77) 


In  this  expression  the  factor  n  is  a  coefficient  of  roughness, 
values  of  which  are  given  in  Table  VI. 

TABLE  VI. — VALUES  OF  n  IN  KUTTER'S  AND  MANNING'S  FORMULAS 


Nature  of  surface 

n 

Planed  and  smoothly  laid  timber  

0  009 

Planed  timber,  not  perfectly  true. 

0  010 

Wood-stave  pipes  

0  Oil 

Smooth  cement  

0  Oil 

Smooth  iron  pipes 

0  Oil 

Rough  timber,  good  brickwork  

0  013 

Slightly  rough  iron  pipes 

0  015 

Rough  brickwork,  cut  stones  

0.015 

Good  rubble  masonry  

0  017 

Tuberculated  iron  pipes 

0  017 

Rough  brick  and  stonework  

0.017 

Smooth  earth  channels  .  . 

0  017 

Coarse  gravel,  well  packed 

0  020 

Large  earth  channels,  good  condition  

0.022 

Small  earth  channels,  good  condition  
Channels  in  fair  condition                                           .  .  . 

0.025 
0  030 

Channels  in  bad  order  with  weeds  etc. 

0  035 

Channels  encumbered  with  drift  

0.045 

In  order  to  save  tedious  computation  when  equation  (77)  is 
used,  various  sets  of  tables  have  been  published  and  also  a  num- 
ber of  graphical  solutions  have  been  devised.1  In  Table  VII 
will  be  found  values  of  c  determined  by  equation  (77).  Inter- 


1  One  of  the  simplest  of  these  is  the  diagram  published  by  Karl  R. 
Kennison. 


130  HYDRAULICS 

TABLE  VII. — VALUES  OF  c  COMPUTED  FROM  KUTTER'S  FORMULA 


Slope 
s 

n 

Hydraulic  mean  depth,  TO,  in  feet 

0.2 

0.4 

0.6 

0.8 

1 

1.5 

2 

3 

4 

6 

8 

10 

15 

0.00005 

0.009 

100 

124 

139 

150 

158 

173 

184 

198 

207 

220 

228 

234 

244 

0.010 

87 

109 

122 

133 

140 

154 

164 

178 

187 

199 

206 

212 

220 

0.011 

77 

97 

109 

119 

126 

139 

148 

161 

170 

182 

189 

195 

205 

0.012 

68 

88 

98 

107 

114 

126 

135 

148 

156 

168 

175 

181 

189 

0.013 

62 

79 

90 

98 

104 

116 

124 

136 

145 

156 

163 

169 

179 

0.015 

51 

66 

76 

83 

89 

99 

107 

118 

126 

137 

144 

149 

158 

0.017 

44 

57 

65 

71 

77 

87 

94 

104 

111 

122 

129 

134 

142 

0.020 

35 

46 

53 

59 

64 

72 

79 

88 

95 

105 

111 

116 

125 

0.025 

26 

35 

41 

46 

49 

57 

62 

71 

77 

85 

91 

96 

104 

0.030 

21 

28 

33 

37 

40 

47 

51 

59 

64 

72 

78 

82 

90 

0.035 

18 

24 

28 

31 

34 

40 

44 

50 

56 

63 

68 

72 

79 

0.0001 

0.009 

112 

136 

149 

158 

166 

178 

187 

198 

206 

215 

221 

226 

233 

0.010 

98 

119 

131 

140 

147 

159 

168 

178 

186 

195 

201 

205 

212 

0.011 

86 

106 

118 

126 

132 

144 

151 

162 

169 

178 

184 

188 

195 

0.012 

76 

95 

105 

114 

120 

130 

138 

149 

155 

164 

170 

174 

181 

0.013 

69 

86 

96 

103 

109 

120 

127 

137 

143 

152 

158 

162 

169 

0.015 

57 

72 

81 

88 

93 

103 

109 

119 

125 

134 

139 

143 

150 

0.017 

48 

62 

70 

76 

81 

89 

96 

104 

111 

119 

124 

128 

135 

0.020 

39 

50 

57 

63 

67 

75 

81 

89 

94 

102 

107 

111 

118 

0.025 

29 

'  38 

44 

48 

52 

59 

64 

71 

76 

84 

88 

92 

98 

0.030 

23 

31 

35 

39 

42 

48 

53 

59 

64 

71 

75 

78 

85 

0.035 

19 

25 

30 

33 

35 

41 

45 

51 

55 

61 

66 

69 

75 

0.0002 

0.009 

121 

143 

155 

164 

170 

181 

188 

200 

205 

213 

218 

222 

228 

0.010 

105 

125 

138 

145 

151 

162 

170 

179 

185 

193 

198 

201 

207 

0.011 

93 

112 

122 

131 

136 

146 

154 

163 

168 

176 

182 

185 

190 

0.012 

83 

100 

111 

118 

123 

133 

140 

149 

155 

162 

167 

170 

176 

0.013 

74 

91 

100 

107 

113 

122 

129 

137 

143 

150 

155 

158 

164 

0.015 

61 

76 

85 

91 

96 

105 

111 

119 

125 

132 

137 

140 

145 

0.017 

52 

65 

73 

79 

83 

91 

97 

105 

111 

117 

122 

125 

131 

0.020 

42 

53 

60 

65 

69 

77 

82 

89 

94 

100 

105 

108 

113 

0.025 

31 

40 

46 

50 

54 

60 

64 

72 

76 

82 

87 

89 

95 

0.030 

25 

32 

37 

41 

44 

49 

54 

59 

63 

69 

73 

76 

82 

0.035 

21 

27 

31 

34 

37 

42 

45 

51 

55 

60 

64 

67 

72 

0.0004 

0.009 

126 

147 

157 

166 

172 

183 

190 

199 

204 

211 

215 

219 

224 

0.010 

110 

129 

140 

148 

154 

164 

170 

179 

184 

191 

196 

199 

203 

0.011 

97 

115 

126 

133 

138 

148 

154 

162 

168 

175 

179 

183 

187 

0.012 

87 

104 

113 

121 

125 

135 

141 

149 

154 

161 

165 

168 

172 

0.013 

78 

94 

103 

110 

115 

124 

130 

138 

142 

149 

153 

157 

162 

0.015 

65 

79 

87 

93 

98 

106 

112 

119 

124 

130 

135 

138 

143 

0.017 

54 

68 

75 

81 

85 

93 

98 

105 

110 

116 

120 

123 

128 

0.020 

44 

55 

62 

67 

70 

78 

83 

89 

94 

99 

104 

107 

110 

0.025 

32 

42 

47 

51 

55 

61 

65 

71 

76 

81 

85 

88 

92 

0.030 

25 

33 

38 

42 

45 

50 

54 

59 

63 

69 

73 

75 

80 

0.035 

21 

27 

31 

35 

37 

42 

45 

51 

55 

60 

64 

66 

70 

0.0010 

0.009 

129 

150 

161 

169 

175 

184 

191 

199 

204 

211 

214 

218 

222 

0.010 

113 

131 

142 

150 

155 

165 

171 

179 

184 

190 

194 

197 

202 

0.011 

99 

117 

127 

134 

139 

149 

155 

163 

168 

174 

178 

181 

186 

0.012 

89 

105 

115 

122 

127 

136 

142 

149 

154 

160 

164 

167 

171 

0.013 

81 

96 

104. 

111 

116 

124 

134 

138 

142 

149 

152 

155 

160 

0.015 

66 

80 

88 

94 

99 

108 

112 

119 

124 

130 

134 

136 

141 

0.017 

57 

69 

76 

82 

86 

93 

98 

105 

110 

116 

120 

122 

127 

0.020 

45 

56 

63 

68 

71 

78 

83 

89 

93 

99 

103 

105 

110 

0.025 

34 

43 

48 

52 

56 

62 

66 

71 

75 

81 

85 

87 

91 

0.030 

27 

34 

39 

42 

45 

50 

54 

59 

63 

68 

72 

74 

78 

0.035 

22 

28 

32 

35 

38 

43 

46 

51 

54 

59 

63 

65 

68 

0.0100 

0.009 

130 

151 

162 

170 

175 

185 

191 

199 

204 

210 

213 

217 

222 

0.010 

114 

133 

143 

151 

156 

165 

171 

179 

184 

190 

193 

196 

200 

0.011 

100 

119 

129 

135 

141 

149 

155 

162 

167 

173 

176 

180 

184 

0.012 

90 

107 

116 

123 

128 

136 

142 

149 

154 

160 

163 

166 

170 

0.013 

81 

98 

106 

112 

117 

125 

130 

138 

142 

148 

151 

154 

159 

0.015 

66 

80 

89 

95 

100 

108 

112 

119 

124 

129 

133 

134 

138 

0.017 

57 

70 

77 

82 

87 

94 

99 

105 

109 

115 

118 

121 

126 

0.020 

46 

57 

64 

68 

72 

79 

83 

89 

93 

99 

102 

105 

108 

0.025 

34 

44 

49 

53 

56 

62 

66 

71 

76 

81 

84 

86 

90 

0.030 

27 

35 

39 

43 

45 

51 

55 

59 

63 

68 

71 

74 

77 

0.035 

22 

29 

33 

35 

38 

43 

46 

51 

55 

59 

62 

65 

68 

UNIFORM  FLOW  IN  OPEN  CHANNELS 


131 


mediate  values  may  be  found  by  interpolation  with  as  much 
accuracy  as  the  conditions  warrant. 

Although  c  is  a  function  of  the  slope,  it  will  be  found  that  its 
variation  with  values  of  s  is  not  great.  The  difference  between 
values  of  c  f or  s  =  0.0001  and  s  =  0.0010  is  a  matter  of  from  10 
to  15  per  cent,  at  the  very  most.  The  equation  also  shows  that 
as  s  increases  in  value  its  influence  upon  the  value  of  c  decreases. 
For  all  values  from  s  =  0.0010  up  to  s  =  0.100,  or  even  greater, 
the  change  in  the  value  of  c  is  negligible.  See  Fig.  125,  which  is 
constructed  for  several  values  of  m  when  n  =  0.017.  Similar 
results  would  be  obtained  with  any  other  value  of  n. 


150 
140 


1 130 
8120 


.no 


•100 


70 


10' 


,017 


1.017 


0    .0001          .0003          .0005  .0010 

Values  of  8 

FIG.  125. — Relation  between  c  and  s. 


.0015 


From  the  range  of  experiments  upon  which  it  is  based,  Kut- 
ter's  formula  would  appear  to  be  applicable  for  values  of  m  up  to 
10  ft.,  for  velocities  up  to  10  ft.  per  sec.,  and  for  slopes  greater 
than  s  =  0.0001.  Outside  of  these  limits  reliable  data  is  lacking 
and  Kutter's  formula  should  be  used  with  caution. 

97.  Manning's  Formula  for  c. — Since  the  value  of  c  is  affected 
by  the  slope  to  a  very  small  extent  only,  as  shown  by  Fig.  125,  both 
Manning  and  Bazin  disregard  its  influence  altogether  and  pro- 
pose formulas  in  which  c  is  independent  of  s.  This  results  in 
much  simpler  expressions.  The  formula  of  Manning  gives  almost 
the  same  values  as  that  of  Kutter  and  is  probably  as  accurate  as 
the  circumstances  warrant.  It  is 


1.49 

c  =  -    -  m 
n 


0.17 


(78) 


132  HYDRAULICS 

in  which  n  is  the  same  as  in  Kutter's  formula.  Values  of  n  are 
given  in  Table  VI. 

Equation  (78)  gives  a  better  idea  of  the  way  c  varies  with  n 
and  m  than  can  be  obtained  from  an  inspection  of  equation  (77). 
Since  these  two  equations  give  values  of  c  which  are  approxi- 
mately equal  to  each  other,  it  follows  that  in  Kutter's  formula  c 
varies  approximately  inversely  as  n  and  directly  as  m0-17. 

For  practical  use  it  is  better  to  compute  V  directly,  rather  than 
to  determine  c  separately  by  Manning's  formula.  Substituting 
the  value  of  c  given  by  equation  (78)  in  equation  (68)  we  obtain 

V  =  —  Vm*  Vs.  (79) 

tv 

Values  of  m"  may  be  found  on  page  263. 

98.  Bazin's  Formula  for  c.  —  For  small  artificial  channels  with 
values  of  m  less  than  3  ft.  and  for  velocities  of  flow  of  not  more 
than  4  ft.  per  sec.,  the  formula  of  Bazin  is  considered  excellent. 
It  is 


'  / 

V  m 

where  n'  depends  upon  the  roughness  of  the  surface.  It  will  be 
noted  that  this  formula  does  not  give  c  as  a  function  of  s.  In 
Table  VIII  will  be  found  values  of  n'  for  use  in  equation  (80). 

TABLE  VIII.  —  VALUES  OF  n'  IN  BAZIN'S  FORMULA 

Nature  of  surface  n' 

Smooth  cement,  or  planed  wood  ........................   0  .  109 

Rough  lumber,  cut  stone,  and  brickwork  ................   0.290 

Rubble  masonry  ......................................   0.  833 

Earth  channels  with  regular  surfaces  .....................    1  .  54 

Ordinary  earth  channels  ................................   2  .  35 

Rough  earth  channels  with  boulders  and  weed-grown  sides.  3.  17 

99.  Construction  of  Open  Channels.  —  Inspection  of  the 
expression  V  =  c\/ms  shows  that,  for  a  given  slope  and  degree  of 
roughness,  the  velocity  increases  as  m  increases.  This  is  also 
accentuated  by  the  fact  that  the  value  of  c  also  increases  as  m 
increases  or,  as  shown  by  equation  (79),  V  varies  as  m**«  There- 
fore for  a  given  area  of  water  cross-section  the  rate  of  discharge 
will  be  a  maximum  when  m  is  a  maximum.  Or  for  a  given  rate 
of  discharge  the  cross-section  area  will  be  a  minimum  when  the 
design  is  such  as  to  make  m  a  maximum. 


UNIFORM  FLOW  IN  OPEN  CHANNELS 


133 


From  equation  (56)  it  may  be  seen  that  the  value  of  m  will  be  a 
maximum  for  a  given  area  when  the  length  of  the  wetted  perime- 
ter is  a  minimum  Now  of  all  geometric  figures,  the  circle  has 
the  least  perimeter  for  a  given  area.  Hence  a  semicircular  open 
channel  will  discharge  more  water  than  one  of  any  other  shape, 
assuming  that  the  area,  slope,  and  roughness  of  surface  are  the 
same.  Semicircular  open  channels  are  often  built  of  pressed 


From  a  photograph  by  the  author. 

FIG.  126. — Open  channel  with    steep    slope.     Future  power  site  on  Los 
Angeles  Aqueduct.     Drop  in  elevation  "=  524  ft. 

steel  and  other  forms  of  metal,  but  for  other  types  of  construc- 
tion such  a  shape  is  impractical.  (The  open  channel  in  Fig.  126 
has  a  semicircular  steel  lining.  This  channel  has  a  wooden 
covering  over  it  but  that  does  not  convert  it  into  a  pressure 
conduit.) 

For  wooden  flumes  the  rectangular  shape  is  usually  used.     Of 
all  rectangles  the  square  has  the  least  perimeter  in  proportion  to 


134 


HYDRAULICS 


area  and  hence  for  an  open  channel  the  depth  of  the  water  should 
be  half  the  width. 

Canals  excavated  in  earth  must  have  a  trapezoidal  section,  and 
of  all  trapezoids  the  half  hexagon  will  have  the  largest  value  of  m. 
But  the  angle  6  cannot  always  be  made  equal  to  60°  for  other 
reasons.  The  slope  of  the  sides  must  be  such  that  the  angle  6  is 


FIG.  127. 


From  a  photograph  by  the  author. 

FIG.  128. — Unlined  canal  with  steep  banks. 


In  the  Sierra  Nevada  Mts. 


less  than  the  " angle  of  repose"  of  the  material  of  which  the  banks 
are  composed,  otherwise  the  latter  will  cave  in.  In  Fig.  128  the 
angle  0  is  made  much  greater  than  60°  in  order  to  save  a  consider- 
able amount  of  excavation  in  a  deep  cut,  the  firm  character  of  the 
soil  permitting  such  steep  sides. 


UNIFORM  FLOW  IN  OPEN  CHANNELS 


135 


Whatever  the  value  of  the  angle,  it  will  be  found  that  the  best 
proportions  will  be  obtained  when  the  sides  are  tangent  to  a 
semicircle  whose  center  lies  in  the  water  surface. 

But  other  forms  of  cross-section  are  often  used  either  because 
they  have  certain  advantages  in  construction  or  are  desirable 
from  other  standpoints.  Thus  oval  or  egg-shaped  sections  are 
common  for  sewers  and  similar  channels  where  there  may  be 
large  fluctuations  in  the  rate  of  discharge.  It  is  desirable  that 
the  velocity,  when  a  small  quantity  is  flowing,  be  kept  high 
enough  to  prevent  the  deposit  of  sediment,  and  when  the  conduit 
is  full  the  velocity  should  not  be  too  high  on  account  of  wearing 
the  lining  of  the  channel. 

100.  Non-uniform  Flow  in  Open  Channels. — As  a  rule  uniform 
flow  is  found  only  in  artificial  channels  of  regular  shape  and  slope, 
although  even  under  these  conditions  the  flow  for  some  distance 
may  be  non-uniform  as  shown  in  Fig.  123.  But  with  a  natural 
stream  the  slope  of  the  bed  and  the  form  and  size  of  the  cross- 
section  usually  vary  to  such  an  extent  that  true  uniform  flow  is 
rare.  Hence  the  application  of  the  equations  given  in  this  chap- 
ter to  natural  streams  can  be  expected  to  yield  results  which  are 


New  Water  Surface 


FIG.  129. 

only  rough  approximations  to  the  truth.  In  order  to  apply  these 
equations  at  all  it  is  necessary  to  divide  the  stream  up  into 
lengths  within  which  the  conditions  are  approximately  the  same. 
A  satisfactory  and  reliable  treatment  of  the  problem  of  non-uni- 
form flow  in  open  channels  is  lacking. 

In  the  case  of  pressure  conduits  we  have  dealt  with  uniform 
and  non-uniform  flow  without  drawing  any  distinction  between 
them.  This  is  possible  because  in  a  closed  pipe  the  area  of  the 
water  cross-section,  and  hence  the  velocity,  is  fixed  at  every  point. 
But  in  an  open  channel  these  conditions  are  unknown  and  the 
stream  adjusts  itself  to  the  size  of  cross-section  that  the  slope  of 
the  hydraulic  gradient  requires. 

In  either  artificial  or  natural  streams  non-uniform  flow  may  be 
produced  in  a  variety  of  ways,  each  one  of  which  leads  to  a  differ- 


136 


HYDRAULICS 


ent  hydraulic  phenomenon.  The  one  case  that  is  of  the  most 
practical  importance  is  where  a  dam  or  other  obstruction  is 
placed  across  a  flowing  stream  (see  Fig.  129).  It  is  desirable  to 
determine  how  far  up-stream  the  "backwater"  created  by  this 
dam  will  extend,  or  at  a  given  point,  how  far  the  water  level  will 
be  raised. 

A  mathematical  treatment  of  this  case  is  given  in  various 
books,  but  it  rests  upon  assumptions  of  conditions  which  are 
rarely  fulfilled  in  natural  streams,  so  that  its  accuracy  for  most 
practical  cases  is  doubtful.1  In  view  of  the  fact  that  it  is  only  an 
approximation,  it  seems  fully  as  satisfactory  to  apply  the  simple 
equation,  V  =  c\/ms,  to  the  case.  In  order  to  do  this  the 
stream  must  be  divided  up  into  various  lengths  within  which  the 
flow  may  be  assumed  to  be  fairly  uniform.  Then  for  each 
length  an  average  value  of  m  and  s  should  be  used  and  the  solu- 
tion completed  for  that  length.  Of  course  the  solution  for  one 
portion  must  be  consistent  with  that  for  other  portions,  that  is, 
the  same  rate  of  discharge  must  exist  for  all  of  them. 

101.  Stream  Gaging. — The  determination  of  the  rate  of  dis- 
charge of  a  stream  for  any  given  depth  of  water  is  termed  stream 
gaging.  It  may  be  seen  that  the  rate  of  discharge  of  a  stream 
could  be  computed  from  the  formula,  V  =  c\/^ns,  if  the  flow  is 
uniform  and  the  cross-section  area,  the  hydraulic  mean  depth, 
and  the  slope  of  the  water  surface  are  known.  But  a  more  accu- 
rate determination  of  dis- 
charge can  be  made  by  meas- 
uring the  velocity  directly, 
by  passing  the  water  over  a 
weir,  or  by  other  means. 

In  a  straight  portion  of  an 
artificial  channel  the  velocity 
might  vary  as  shown  in  Fig. 
130.  These  curves  are  veloc- 
ity contours  or  curves  of 
equal  velocity.  Within  the  area  enclosed  by  the  curve  the 
velocity  is  higher  than  that  of  a  point  on  the  curve.  Outside 
the  enclosed  area  the  velocity  is  less  than  that  on  the  curve. 

1  The  formula  derived  by  calculus  to  fit  this  case  is  based  upon  the 
assumptions  that  the  slope  of  the  bed  is  uniform,  that  the  form  of  the 
water  cross-section  is  uniform  except  that  its  depth  varies,  and  that  the 
stream  is  very  broad  as  compared  to  its  depth. 


FIG.  130. 


UNIFORM  FLOW  IN  OPEN  CHANNELS          137 

It  may  be  seen  that  the  velocity  of  the  water  varies  from  side  to 
side  and  from  top  to  bottom.  If  there  is  a  bend  in  the  channel, 
or  if  the  bed  is  irregular,  as  in  natural  streams,  these  velocity 
curves  are  often  very  irregular  and  distorted  from  the  forms 
shown  here.  It  is,  therefore,  necessary  to  determine  the 
velocity  at  a  number  of  different  points  across  the  section  of  the 
stream. 

The  instrument  that  is  commonly  used  for  this  purpose  is  the 
current  meter,  described  in  Art.  70.  In  using  the  current  meter 
or  any  other  device  it  is  cus- 
tomary to  divide  the  stream 
up  into  sections  as  in  Fig.  131 
and  to  determine  the  contour 
of  the  bed,  so  that  the  area 

may  be  computed.      If  then  pIG   131 

the   average   velocity   is   de- 
termined for  some  section  such  as  A  BCD  the  discharge  through 
this  section  will  be  the  product  of  this  velocity  and  the  area 
A  BCD.     The  sum  of  all  such  partial  discharges  gives  the  total 
rate  of  discharge  of  the  entire  stream. 

In  finding  the  average  velocity  in  the  area  A  BCD  it  is  cus- 
tomary to  take  it  as  the  average  of  the  velocity  measured  in  the 
line  AB  and  the  velocity  measured  .in  the  line  CD.  But,  as 
shown  in  Fig.  132,  the  velocity  varies  from  A  to  B  or  from  C  to 
__^^ *<==  A  and  hence  we  should  determine  the 

average  velocity  in   each  vertical  line. 

This  might  be  done  by  taking  a  num- 

— i         -,—     ber    of    observations    so    that     curves 

j  Average  velocity^      similar  to  that  in   Fig.   132  could  be 

_^=          plotted.     But  a  study  of  a  number  of 

~~^r-  such  curves  has  shown  that  in  general 

the  average  velocity  in  a  vertical  line 


>fe 

1 


FlG   132  ig  found  at  about  0.6  the  depth.     Hence 

if  the  current   meter   be  set  at  that 

depth,  the  velocity  determined  by  it  may  be  assumed  to  be  the 
mean  velocity.  Of  course  this  is  only  an  approximation.  To 
insure  a  higher  degree  of  accuracy  than  a  single  observation  could 
give,  measurements  are  often  taken  at  0.2  the  depth  and  0.8  the 
depth.  The  mean  of  these  two  values  will  be  approximately  the 
average  velocity.  Thus,  in  an  actual  stream  gaging,  observa- 
tions would  be  made  at  the  points  indicated  by  the  circles  in 


138 


HYDRAULICS 


Fig.  131.  Further  details  of  this  topic  are  not  within  the  scope 
of  this  text.1 

Sometimes  floats  are  used  but  such  procedure  is  less  accurate. 
However  they  are  often  applicable  when  other  methods  are  not 
feasible,  such  as  during  floods.  If  surface  floats  are  used,  the 
average  velocity  may  ordinarily  be  taken  as  about  0.9  that  of  the 
surface  velocity.  But  the  velocity  at  the  surface  is  greatly 
affected  by  the  wind. 

102.  Rating  Curve. — If  a  natural  stream  is  to  be  used  for  water 
supply  or  power  purposes,  it  is  necessary  to  determine  the 
amount  of  water  it  can  be  depended  upon  to  furnish.  Since  the 
flow  will  usually  be  subject  to  wide  fluctuations  during  a  long 
period  of  time  it  is  necessary  to  make  an  extended  series  of 
observations  upon  it. 


Rate  of  Discharge 
FIG.  133. — Rating  curve. 

The  level  of  the  surface  of  the  water  in  a  stream  is  called  the 
gage-height,  and  may  be  measured  above  any  arbitrary  point. 
Thus  the  gage-height  does  not  necessarily  coincide  with  the 
depth  of  the  stream. 

It  is  apparent  that  for  a  given  stream,  the  rate  of  discharge  will 
be  a  function  of  the  gage-height.  If  the  rate  of  discharge  of 
the  stream  be  determined  for  several  gage-heights  a  curve,  such 
as  in  Fig.  133,  may  be  constructed.  This  curve  is  called  the 
rating  curve,  and  from  it  the  value  of  q  for  any  height  of  water 
can  be  obtained. 

Thus  in  making  a  study  of  the  stream  it  is  necessary  to  make 
only  a  record  of  the  gage-heights.  From  the  rating  curve  the 
quantity  of  flow  can  then  be  determined.  This  gage-height 
might  simply  be  read  and  recorded  once"  a  day  by  an  observer,  or 
by  means  of  a  float  and  clockwork  a  continuous  record  could  be 

1  See  Hoyt  and  Grover,  "River  Discharge." 


UNIFORM  FLOW  IN  OPEN  CHANNELS 


139 


obtained  which  would  show  all  the  variations  in  the  flow.     Fig. 
87  shows  such  a  gaging  station. 


103.  PROBLEMS 

1.  A  circular  conduit  of  smooth  cement  is  exactly  half  full  of  water.     The 
diameter  is  4  ft.  and  the  slope  is  1  ft.  per  10,000  ft.     Compute  the  rate  of 
discharge  by  the  formulas  of  Kutter,  Manning,  and  Bazin. 

Am.  q  =  8.38,  8.51,  and  8.92  sec.  ft.  respectively. 

2.  A  rectangular  flume  of  timber  slopes  1  ft.  per  1,000  ft.     Compute  the 
rate  of  discharge  if  the  width  is  6  ft.  and  the  depth  of  water  3  ft. 

Ans.   114  sec.  ft. 

3.  What  would  be  the  rate  of  discharge  in  problem  (2)  if  the  width  were  3 
ft.  and  the  depth  of  water  6  ft.  ?     Which  of  the  two  forms  would  require  less 
lumber? 

4.  A   rectangular  channel  of  rubble  ma- 
sonry is  6  ft.  wide,  the  depth  of  water  is  3 
ft.,  and  the  slope  of  1  ft.  per  1,000  ft.    Com- 
pute the  rate  of  discharge  and  compare  with 
that  in  problem  (2). 

Ans.  65  sec.  ft. 

6.  A  semicircular  channel  of  rubble  ma- 
sonry with  a  slope  of  1  ft.  per  1,000  ft.  will 
give  what  discharge  when  flowing  full  if  its 
diameter  is  6.55  ft.?  Compare  the  cross- 
section  areas  and  amounts  of  lining  required 
with  that  in  problem  (4). 
Ans.  65  sec.  ft. 

6.  A  circular  conduit  of  concrete  (n  =  0.012)  is  10  ft.  in  diameter  and 
slopes  1.6  ft.  per  1,000  ft.  (See  Fig.  134).  The  following  table  gives  values  of 
wetted  perimeter  and  area  of  water  cross-section  for  various  depths  of  water 
in  the  conduit.  Find  values  of  V  and  q  for  the  various  depths  in  the  table. 
What  value  of  y  gives  the  highest  velocity?  What  value  of  y  gives  the 
highest  rate  of  discharge? 


FIG.  134. 


Depth,  y 

Wetted  perimeter 

Area,  F 

m 

vs 

c 

V 

a 

1.0 

6.44 

4.09 

3.0 

11.59 

19.82 

5.0 

15.71 

39.27 

8.0 

22.14 

67.36 

9.0 

24.98 

74.45 

9.5 

26.91 

77.07 

10.0 

31.42 

78.54 

7.  The  amount  of  water  to  be  carried  by  a  canal  excavated  in  firm  gravel 
is  370  sec.  ft.  It  has  side  slopes  of  2  : 1  (horizontal  dimension  is  two  times 
vertical  dimension)  and  the  depth  of  water  is  to  be  5  ft.  or  less  (Fig.  135). 


140 


HYDRAULICS 


If  the  slope  is  2.5  ft.  per  mile,  what  must  be  the  width  at  the  bottom? 
(This  problem  can  best  be  solved  by  trial.) 

8.  Assume  that  the  Cippoletti  weir  in  Fig.  87  is  8  ft.  wide  and  that  the 
depth  of  the  notch  is  1.5  ft.  Construct  a  rating  curve  that  may  be  used  for 
this  gaging  station.  (The  gage  height,  which  in  this  case  is  the  head  on  the 
weir,  is  recorded  by  a  suitable  mechanism  in  the  shelter  shown  in  the  picture.) 
What  would  be  the  nature  of  the  rating  curve  for  values  of  H  greater  than 
1.5  ft.,  where  H  is  measured  from  the  hook  gage  zero? 


FIG.  135. 


CHAPTER  IX 


HYDRODYNAMICS 

104.  Dynamic  Force  Exerted  by  a  Stream. — Whenever  the 
velocity  of  a  stream  of  water  is  changed  either  in  direction  or 
in  magnitude,  a  force  is  required.  By  the  law  of  action  and 
reaction  an  equal  and  opposite  force  is  exerted  by  the  water 
upon  the  body  which  produces  this  change.  This  is  called 
a  dynamic  force  in  order  to  distinguish  it  from  static  forces  due 
to  the  pressure  of  the  water. 


n/l 


AV 


Let  the  resultant  force  exerted  by  any  body  upon  the  water 
be  denoted  by  R  and  its  components  by  Rx  and  Ry.  Let  dR 
be  the  force  exerted  upon  an  elementary  mass  dm  of  cross-section 
area  F  and  length  along  the  path  ds.  (See  Fig.  136.)  Then, 
since  resultant  force  equals  mass  times  acceleration, 


dRx  =  dm 


dVx      wFds  dVx       wFds 


dVx. 


dt  g       dt          gdt 

But  ds/dt  =  V,  and  for  steady  flow  Fv  is  constant. 
wFds  _  wFV  _  wq  _  TF 

9<!t         g        g  "  g 


Hence 


Therefore 


dRx  =  — 

141 


142  HYDRAULICS 

For  steady  'flow  W  is  constant,  thus  we  may  write, 

W  f2  W       I2 


Now  at  point  (1)  the  value  of  F*  is  F]  cos  «i  and  at  (2)  it  is  F2 
cos  az.     Inserting  these  limits  we  have 

W 

Rx  =  —  (F2  cos  «2  —  Fi  cos  «i) 
Q 

If  P  indicates  the  value  of  the  force  exerted  by  the  water, 
which  is  equal  and  opposite  to  R,  we  shall  have 

W 

Px  =  —  (Fi  cos  ai  -  F2  cos  «2).  (81) 

In  similar  manner  the  y  component  of  P  will  be 

W 

Py  =  —  (Fi  sin  on  —  F2  sin  «2).  (82) 

From  Fig.  136  it  may  be  seen  that  Fi  cos  on  —  F2  cos  «2  = 

— AFje  and  Fi  sin  a\  —  F2"sin  «2  =  — AFV. 

Since 

P  =VPX2  +  PV2  and  AF  =  VAF*2  +  AFtf2. 
the  value  of  the  resultant  force  is 

W 

P  =  —  AF.  (83) 

The  direction  of  R  will  be  the  same  as  that  of  AF  and  the  direction 

of  P  will  be  opposite  to  it.     Note 
that  AF  is  the  vector  difference 
between  FI  and  F2. 
I  If  the  angle  between  Fi  and 

F2  be  denoted  by  0  as  in  Fig. 
137    (0  =  «2  -  ai),  and  the  x 

axis  be  taken  parallel  to  Fi,  so  that  cos  a\  =  1.0,  we  may  write 

px  =  ~  (V,  -  F2  cos  0)  (84) 

and 

Pv  = F2  sin  0.  (85) 

These  forms  will  often  be  more  convenient. 

EXAMPLES 

1.  In  Fig.  137  assume  that  6  =  60°,  and  that  the  stream  striking  the 
body  is  a  jet  2  in.  in  diameter  with  a  velocity  of  100  ft.  per  sec.  If  the 
frictional  loss  is  such  as  to  reduce  the  velocity  of  the  stream  leaving  the 


HYDRODYNAMICS  143 

body  to  80  ft.  per  sec.,  find:  (a)  the  component  of  the  force  in  same  direc- 
tion as  the  jet,  (6)  the  component  of  the  force  normal  to  the  jet,  (c)  the  magni- 
tude and  direction  of  the  resultant  force  exerted  by  the  water. 

Ans.   (a)  254  Ib.     (6)  293  Ib.     (c)  388  Ib.  at  49°  08'  with  direction  of  jet. 

2.  Suppose  the  jet  in  problem  (1)  struck  a  flat  plate  normally,  what 
would  be  the  value  of  the  force  exerted  upon  the  plate? 

Ans.  423  Ib. 

3.  Suppose  the  jet  in  problem  (1)  were  completely  reversed  in  direction, 
or  that  6  =  180°.     If  F2  were  100  ft.  per  sec.,  what  would  be  the  component 
of  the  force  in  the  same  direction  as  the  jet?     (Compare  with  problem  (2).) 
What  would  be  the  component  normal  to  the  direction  of  the  jet? 

Ans.  846  Ib 

4.  Suppose  that  in  problem  (3)  the  value  of  V*  were  reduced  to  80  ft. 
per  sec.  as  in  problem  (1).     What  would  be  the  value  of  the  force  exerted? 
(Compare  with  problem  (3).) 

Ans.  761  Ib. 

105.  Force  Exerted  upon  Pipe. — When  a  flowing  stream  is 
confined  there  may  be  static  forces  due  to  pressure  as  well  as 
dynamic  forces  due  to 
changes  in  velocity.  Con- 
sider the  water  to  be  flowing 
to  the  right  in  Fig.  138.  -^> 
Since  the  velocity  is  increased 

from  Vi  to  Vtj  the  dynamic  pIQ   138 

force  exerted  upon  the  water, 
according  to  equation  (83),  will  be 

R  =  j  (V*  -  70. 

This  force,  producing  the  acceleration  of  the  water,  must  be  the 
resultant  of  all  the  forces  acting.  The  real  forces  acting  upon 
the  volume  of  water  shown  are  the  pressures  upon  the  two  ends 
p'iFi  and  p'zFz  exerted  by  the  rest  of  the  water,  and  the  force 
N  exerted  by  the  pipe  walls.1  If  there  were  no  friction  this  force 
would  be  normal  to  the  walls,  but  actually  it  will  be  inclined 
somewhat  from  the  normal  because  it  must  have  a  frictional 
component.  Let  the  component  of  N  parallel  to  the  axis  of 
the  pipe  be  denoted  by  Nx.  It  may  be  seen  that  R  must  be  in 
the  same  direction  as  Vi  and  72  in  Fig.  138.  Hence  the  sum 

1  The  N  shown  in  Fig.  138  represents  the  force  for  an  element  only.  For 
a  pipe  of  circular  cross-section  the  resultant  force  exerted  by  the  walls  must 
be  axial. 


144 


HYDRAULICS 


of  all  the  forces  parallel  to  the  axis  of  the  pipe  must  equal  R. 
Therefore 

R  =  p'iFi  -  pf2F2  -  N,. 


Inserting  the  value  of  R  given  above,  it  follows  that 


N,  = 


-  pf2F2  -        (F2  - 


(86) 


It  must  be  remembered  that  Nx  is  assumed  to  be  the  axial  com- 
ponent of  the  force  exerted  upon  the  water  by  the  conical  por- 
tion of  pipe.  The  force  exerted  by  the  water  upon  the  pipe 
is  equal  and  opposite  to  this.  That  is,  its  magnitude  is  given  by 
equation  (86)  but  it  acts  toward  the  right. 

If  the  velocity  of  the  water  in  a 
closed  passage  undergoes  a  change 
in  its  direction,  as  in  the  pipe  bend 
shown  in  Fig.  139,  the  procedure 
would  be  similar  to  that  in  the 
preceding  case.  The  forces  acting 
on  the  water  in  the  bend  are  the 
pressures  p'iFi  and  p'2F2  and  the 
pressure  exerted  by  the  walls  of 
the  pipe,  designated  by  N.  By 
equation  (83)  the  resultant  force  acting  upon  this  volume  of 

W 

water  will  be  R  =  — AF,  but   R  is  the  resultant  of  the  three 

y 

forces  just  mentioned.  Since  these  are  vector  quantities  not  in 
the  same  straight  line,  it  will  be  better  to  take  x  and  y  com- 
ponents. Thus  we  should  write 


FIG.  139. 


W 

Rx  =  —  (F2  cos  0  -  Fi)  =  p'iFi  -  p'2F2  cos  0  -  Nx 


and 


W 

Ry  =  —  V2  gin 
y 


-  p'2F2  sin  6  +  Nv. 


Solving  these  equation  we  find  that 


W 
Nx  =  p\Fi  -  p'2F2  cos  6  +  — 


-  F2  cos  0) 


and 


JVy  = 


sn 


W 

H  --  F2  sin  8. 

\y 


(87) 


(88) 


But  again  N  represents  the  force  exerted  by  the  pipe  bend  upon 


HYDRODYNAMICS 


145 


the  water.     The  force  exerted  by  the  water  upon  the  bend  will  be 
equal  and  opposite  to  this. 

It  may  be  seen  that  these  forces  tend  to  move  the  portion  of 
pipe  considered.  Hence  a  pipe  should  be  ''anchored"  where 
such  changes  in  velocity  occur. 


EXAMPLES 

1.  On  the  end  of  a  6-in.  pipe  is  a  nozzle  which  discharges  a  jet  2  in.  in 
diameter.     The  pressure  in  the  pipe  is  55  Ib.  per  sq.  in.  and  the  pipe  velocity 
is  10  ft.  per  seci     The  jet  is  discharged  into  the  air.     (a)  What  is  the  re- 
sultant force  acting  on  the  water    within  the  nozzle?     (6)   What  is  the 
axial  component  of  the  force  exerted  on  the  nozzle? 

Ans.  (a)  304  Ib.     (6)  1,250  Ib. 

2.  Water  under  a  pressure  of  40  Ib.  per  sq.  in.  flows  with  a  velocity  of 
8  ft.  per  sec.  through  a  right-angle  bend  having  a  uniform  diameter  of  12  in. 
(a)  What  is  the  resultant  force  acting  on  the  water?     (6)  What  is  the  total 
force  exerted  on  the  bend? 

Ans.  (a)  137.8  Ib.     (6)  6,530  Ib. 

106.  Theory  of  Pitot  Tube. — The  Pitot  tube  has  been  briefly 
described  in  Art.  69  and  illustrated  in  Fig.  95.     We  shall  now 


consider  the  dynamic  action  of  the  water  upon  it.  In  Fig. 
140  a  Pitot  tube  is  placed  with  its  opening  facing  upstream, 
the  velocity  of  the  water  being  denoted  by  V.  The  dotted  lines 
in  the  figure  are  intended  to  represent  an  imaginary  cylinder 
of  cross-section  area  F  equal  to  that  of  the  mouth  of  the  tube 
and  extending  to  point  (1)  as  far  up  stream  as  the  influence  of 
the  tube  is  felt.1 

1  The  method  of  derivation  of  the  Pitot  tube  formula  given  here,  as  well 
as  some  interesting  experimental  results,  will  be  found  in  a  paper  by  L.  F. 
Moody,  "Measurement  of  the  Velocity  of  Flowing  Water,"  Proc.  of  the 
Engineers'  Soc.  of  West.  Penn.,  vol.  30,  page  279  (1914). 
10 


tr* 


146  HYDRAULICS 

At  point  (1)  the  water  within  this  cylinder  has  a  velocity  V 
and,  as  it  approaches  the  Pitot  tube,  its  velocity  continually 
decreases  until  it  becomes  zero  at  point  (2).  But  if  water 
flows  into  this  cylinder,  bounded  by  the  dotted  lines,  it  must 
also  flow  out.  It  does  this  along  the  sides,  for  since  the  area  F 
is  constant  while  the  velocity  is  a  decreasing  quantity,  it  follows 
that  q  (within  the  cylinder)  must  become  less  as  the  Pitot  tube 
is  approached.  The  conditions  for  a  certain  mass  of  water  are 
therefore  as  shown  in  Fig.  140 (a).  As  the  velocity  of  the  water 
decreases  the  cross-section  area  must  increase  for  the  same  value 
of  q.  Referring  to  Fig.  140 (a),  consider  that  water  flows  into 
this  portion  of  the  stream  with  a  velocity  V  and  leaves  it  with 
a  velocity  V  —  AF'.  If  the  cross-section  area  of  the  stream 
entering  the  section  is  F,  we  have  W  =  wq  =  wFV.  If  the  two 
faces  of  this  volume  be  taken  at  an  infinitesimal  distance  apart 
the  velocity  will  decrease  by  an  amount  —  dV,  hence  the  dynamic 
force  exerted  upon  this  small  mass  of  flowing  water  will  be 

dR  =  -  —  dV'  =  -  —  V'dV. 
Q  9 

The  value  of  V  varies  from  V  at  point  (1)  to  0  at  point  (2). 
Since  F  is  constant  we  may  write 


g       2 

The  dynamic  force  exerted  by  the  flowing  water  upon  the  body 
of  still  water  within  the  Pitot  tube  is  equal  and  opposite  to  R. 
If  the  force  be  represented  by  P,  we  have 

P  =  wF  ~  (89) 

This  is  the  value  of  the  total  force  distributed  over  the  area  F. 
The  intensity  of  pressure  is  pf  =  wV2/2g,  or  dividing  by  w  we 
have  intensity  of  pressure  in  feet  of  water  so  that 

V2 
*  -  2  '  (90) 

That  this  is  true  has  been  amply  demonstrated  by  experimental 
evidence.  If  the  water  is  under  pressure,  the  Pitot  tube  will 
read  the  sum  of  the  pressure  head  and  the  dynamic  head,  given 
by  equation  (90).  It  is  therefore  necessary  to  determine  the 


HYDRODYNAMICS  147 

pressure  head  separately  or  else  use  a  differential  manometer, 
one  side  of  which  shall  be  connected  to  the  Pitot  tube  and 
the  other  to  a  piezometer  tube.  The  chief  source  of  error  in 
the  use  of  the  Pitot  tube  lies  in  the  measurement  of  the  pressure. 

If  the  Pitot  tube  is  used  in  a  pipe,  the  pressure  reading  should 
be  taken  by  a  piezometer  tube  which  does  not  project  within 
the  walls  of  the  pipe  and  which  is  at  right  angles  to  it,  as  in  the 
second  tube  of  Fig.  95.  If  it  is  necessary,  for  some  reason,  to 
have  the  tube  project  into  the  stream,  a  correct  reading  may  be 
obtained  if  the  piezometer  orifice  is  made  in  a  flat  plate,  the  plane 
of  which  may  be  parallel  to  the  stream  lines,  as  is  shown  by  the 
third  tube  in  Fig.  95.  Or  the  orifice  may  be  made  in  the  side  of 
a  smooth  tube,  whose  axis  is  parallel  to  the  stream  lines  and 
whose  closed  upstream  end  is  pointed  so  as  to  diminish  eddy 
disturbances.1 

107.  Water  Hammer  and  Surges  in  Unsteady  Flow.  —  In  all 
the  rest  of  this  book  the  treatment  is  restricted  to  cases  of 
steady  flow,  but  in  the  present  article  a  brief  description  will  be 

1  The  Pitot  tube  formula  has  often  been  derived  by  an  incorrect  ap- 
plication of  the  principles  of  Art.  104.  If  a  jet  of  water  with  cross-section 
area  F  impinges  normally  upon  a  flat  plate,  the  dynamic  force  will  be 


g  g  g 

This  is  twice  the  value  given  by  equation  (89),  and  dividing  this  by  the  area 
of  the  Pitot  tube  orifice,  which  is  also  assumed  equal  to  F,  the  intensity  of 

V2 
pressure  in  feet  of  water  is  apparently  h  =  —.     But  this  reasoning  is  in- 

correct; for  a  flat  plate  of  an  area  the  same  as  that  of  a  jet  would  not  be  able 

to  deflect  all  the  water  through  an  angle  of 

90°.     Experiment  shows  that  the  dynamic 

pressure  exerted  by  a  circular  jet  is  distrib- 

uted over  a  circular  area  whose  diameter  is 

at  least  twice  that  of  the  jet.     Therefore  if 

the  entire  stream  of  water  is  to  be  deflected 

through  an  angle  of  90°  the  area  of    the 

plate  must  be  at  least  four  times  that  of 

the  jet.     Dividing  P  by  4F  we  should  have 

the  average  intensity  of  pressure  to  be  pf  = 

wV2/4:g.     It  is  found  experimentally  that 

the  maximum  intensity  of  pressure  at  the  FIG.  140(6). 

center  of  the  plate  in  feet  of  water  is  V2/2g, 

and  that  this  pressure  diminishes  in  intensity  as  the  outer  margins  of  the 

area  in  question  are  approached,  as  shown  in  Fig.  140(6). 

See  "Pitot  Tube  Formulas—  Facts  and  Fallacies"  by  B.  F.  Groat,  Proc. 
of  Engineers'  Soc.  of  West.  Penn.,  vol.  30,  page  324  (1914). 


148 


HYDRAULICS 


given  of  the  problems  of  unsteady  flow  that  are  of  the  most 
practical  importance.  An  adequate  mathematical  treatment 
of  unsteady  flow  would  occupy  too  much  space  to  warrant  its 
inclusion  here  and  no  attempt  will  be  made  to  do  more  than 
record  some  accepted  results. 

In  the  event  of  a  valve  at  C  in  Fig.  141  (a)  being  rapidly  closed 
in  a  short  interval  of  time  At,  the  velocity  of  the  water  in  the 
pipe  will  be  abruptly  reduced  to  zero.  But  in  so  doing  there 
will  be  a  considerable  rise  in  pressure  within  the  pipe,  which  may 
be  much  greater  than  any  static  pressure  that  could  possibly 
exist  in  the  given  pipe.  This  high  pressure  lasts  for  an  instant 
only,  and  then  follows  a  periodic  fluctuation  of  pressure  which 
finally  dies  out,  if  the  pipe  does  not  burst  in  the  meantime. 
This  is  known  as  water  hammer. 

What  happens  is  that  the  lamina  of  water  next  to  the  valve 
at  C  is  brought  to  rest  and  is  then  compressed  by  the  rest  of 
the  column  of  water  flowing  up  against  it.  At  the  same  time 


FIG.  141. 

the  walls  of  the  pipe  surrounding  this  lamina  will  be  stretched 
by  the  excess  pressure.  The  next  lamina  of  water  will  be 
brought  to  rest  by  the  first  and  so  on.  It  is  seen  that  the  volume 
of  water  in  the  pipe  does  not  behave  as  a  rigid  body  but  that 
the  phenomena  is  affected  by  the  elasticity  of  the  water  and  the 
pipe.  Thus  the  cessation  of  flow  and  the  increase  of  pressure 
progresses  along  the  pipe  as  a  wave  action.  After  a  short  interval 
of  time  the  volume  of  water  BC  will  have  been  brought  to  rest, 
while  the  water  in  the  length  A  B  will  still  be  flowing  with  its 
initial  velocity,  and  with  its  initial  pressure.  But  the  volume 
of  water  in  BC  will  be  under  a  much  higher  pressure  due  to  the 
compression  it  is  under  and  the  pipe  walls  will  be  stretched.  The 
excess  pressure  DE  is  the  same  for  all  portions  of  the  pipe  and 
is  independent  of  the  length  of  the  pipe. 

Finally  the  pressure  wave  will  have  reached  the  reservoir 
and  the  entire  volume  of  water  will  be  at  rest.     But,  owing  to  its 


HYDRODYNAMICS  149 

compression  as  well  as  the  tension  of  the  pipe,  the  flow  will 
tend  to  start  toward  the  reservoir.  Thus  a  wave  of  rarefaction 
proceeds  from  A  to  C,  and  so  on  until  the  waves  die  out.  If  the 
valve  is  alternately  opened  and  closed  at  just  the  proper  intervals 
of  time  it  is  possible  to  add  one  pressure  wave  on  top  of  another, 
so  that  there  is  no  limit  to  the  maximum  pressure  that  might 
be  attained. 

The  velocity  with  which  this  pressure  wave  progresses  along 
the  pipe  will  be  given  by  the  following  formula:1 


yw  =  4700     -  —,  (91) 

300,000  j- 

where  Vu  —  velocity  of  pressure  wave  in  feet  per  second,  E  = 
modulus  of  elasticity  in  tension  of  the  material  composing  the  pipe 
in  pounds  per  square  inch,  and  d/t  is  the  ratio  of  the  diameter  of 
the  pipe  to  the  thickness  of  the  walls,  which  means  that  both  d 
and  t  must  be  in  the  same  units.  The  values  of  E  for  steel, 
cast  iron,  and  wood  are  about  30,000,000  Ib.  per  sq.  in.,  15,000,- 
000  Ib.  per  sq.  in.,  and  1,500,000  Ib.  per  sq.  in.  respectively. 
For  pipes  of  ordinary  dimensions  the  velocity  of  this  pressure 
wave  will  be  about  3,300  ft.  per  sec.  In  any  event  it  will  be  less 
than  4,700  ft.  per  sec.,  which  is  the  velocity  of  sound  in  water 
or  the  velocity  with  which  a  pressure  wave  would  be  propagated 
in  water  in  a  rigid  pipe.  See  equation  (91). 

The  time  required  for  a  pressure  wave  to  travel  the  length 
of  the  pipe,  or  the  time  that  it  takes  for  the  entire  mass  of 
water  to  be  brought  to  rest  will  be 

T  =  l/Vu  (92) 

The  total  force  exerted  may  be  determined  by  applying 
the  principle  that  force  equals  mass  times  acceleration.  Since 
the  volume  of  water  is  a  non-rigid  body  we  must  deal  with  the 
acceleration  of  the  mass  center.  The  pressure  wave  travels 
at  a  uniform  rate;  hence  the  velocity  of  the  mass  center  is 
uniformly  retarded.  Therefore  the  acceleration  may  be  deter- 
mined by  dividing  the  change  in  velocity  by  the  time  required 
for  the  change  to  occur.  The  velocity  of  all  the  water,  and 

1  The  complete  expression  involves  the  volume  modulus  of  elasticity  of 
the  water,  the  density  of  the  water,  and  the  value  of  g.  Using  average 
values  of  these  quantities  the  above  is  obtained. 


150  HYDRAULICS 

hence  that  of  the  mass  center,  decreases  from  V  to  0  in  the  time 
T.  Thus  we  may  write 

wFl  V      wFl  VVu          nVV» 

P  = TF,  =  ~ r-^  =  u>F 

g    T        g       I  g 

Dividing  by  the  area  F  and  also  by  w  we  obtain  intensity  of 
pressure  in  feet  of  water.  If  this  excess  pressure,  due  to  water 
hammer,  be  denoted  by  pm,  we  have 


It  will  be  noted  that  this  pressure  increase  is  independent  of  the 
length  of  the  pipe  line.1 

The  length  of  the  pipe  line  enters  into  the  problem  in  this 
way.  The  time  required  for  a  pressure  wave  to  make  the 
round  trip  from  the  gate  to  the  reservoir  and  back  is  twice 
the  value  given  by  equation  (92).  It  has  been  found  that 
the  pressure  created  is  independent  of  the  time  of  closure  of 
the  gate  provided  that  it  is  closed  in  less  time  than  it  takes 
for  a  pressure  wave  to  make  the  round  trip.  That  is  the  gate 
must  be  closed  in  less  time  that  21  /Vu.  If  the  time  is  greater 
than  this  the  pressure  is  reduced  in  proportion  as  follows : 

Tr 

P    =    Pmlf7  (94) 

where  Tr  is  the  time  for  a  round  trip  of  the  pressure  wave,  T' 
is  any  time  greater  than  this  and  p  is  the  pressure  that  will  be 
attained  in  such  a  case. 

It  is  seen  that  in  a  short  pipe  line  the  value  of  Tr  is  so  small  that 
it  is  nearly  impossible  to  close  the  gate  quickly  enough  to  pro- 
duce water  hammer  of  maximum  intensity.  In  a  long  pipe 
line  it  is  necessary  to  close  the  gate  slowly  in  order  to. prevent 
this  and  the  longer  the  pipe  line  the  slower  the  gate  must  be 
closed. 

For  the  sake  of  clearness  in  explanation  it  has  been  assumed 
in  the  preceding  discussion  that  the  velocity  of  the  water  has 

1  The  subject  of  water  hammer  has  been  experimentally  investigated  by 
Joukovsky  of  Moscow  on  pipes  of  2-,  4-,  6-,  and  24-in.  diameter  and  with 
lengths  ranging  from  1,050  ft.  to  7,007  ft.  He  found  the  results  to  agree 
with  the  formulas  given.  For  a  resume  of  his  work  see  "  Water  Hammer" 
by  Simin,  Trans.  Amer.  W.  W.  Ass'n,  1904. 


HYDRODYNAMICS  151 

been  reduced  to  zero.  But  the  results  are  true  for  any  reduction 
in  velocity,  it  being  simply  necessary  to  substitute  AF  for  V. 

Water  hammer  may  be  prevented  by  the  use  of  slow  closing 
valves,  or  its  effects  diminished  by  the  use  of  automatic  relief 
valves  which  permit  water  to  escape  when  the  pressure  exceeds 
a  certain  value.  Also  air  chambers  of  suitable  size  provide 
cushions  which  absorb  a  great  portion  of  the  shock.  But  for 
water  power  plants  a  standpipe  or  surge  chamber  such  as  is 
shown  in  Fig.  141(6)  has  certain  marked  advantages. 

In  the  event  of  a  sudden  decrease  in  load  on  a  water  power 
plant  it  would  be  necessary  for  the  governors  to"  rapidly  reduce 
the  amount  of  water  supplied  to  the  wheels,  if  the  speed  of  the 
latter  is  to  be  maintained  constant.  A  surge  chamber  provides 
a  place  into  which  this  excess  water  may  flow  and  thus  avoids 
water  hammer  in  the  supply  pipe.  The  inertia  of  the  mass 
of  water  flowing  down  this  supply  pipe  may  be  such  as  to  carry 
the  water  level  above  the  static  level  and  produce  an  ascending 
hydraulic  gradient.  But  this  excess  pressure  acts  as  a  retarding 
force  on  the  mass  of  water  in  the  pipe  line  and  thus  reduces 
its  velocity.  In  any  event  the  temporary  water  level  in  this 
surge  chamber  will  be  higher  than  the  normal  value  and  hence 
it  will  reduce  the  velocity  of  flow  too  much.  The  result  will  be 
that  there  will  be  fluctuations  of  velocity  in  the  pipe  line 
accompanied  by  "surges"  of  the  water  level  in  the  chamber 
until  a  condition  of  equilibrium  is  finally  reached.  The  phe- 
nomenon is  very  similar  to  that  of  water  hammer  as  there  are 
periodic  alternations  in  pressure  and  velocity,  but  the  pressure 
variations  are  much  less  severe. 

The  surge  chamber  fulfills  another  valuable  function  in  that 
it  not  only  takes  care  of  excess  water  in  case  of  a  sudden  re- 
duction of  flow  but  it  also  provides  a  source  of  water  supply  in 
the  event  of  a  sudden  demand.  When  the  load  on  the  plant 
increases  it  is  necessary  to  supply  more  water  to  the  wheels  at 
once.  If  the  pipe  line  is  long  it  may  take  some  time  to  accelerate 
the  entire  mass  of  water  and  in  the  meantime  the  head  at  the 
plant  has  dropped  considerably  in  order  to  provide  an  ac- 
celerating force.  But  the  surge  chamber  permits  a  certain 
amount  of  water  to  flow  out  during  that  period.  To  be  sure 
enough  flows  out  so  that  the  hydraulic  gradient  drops  below 
its  normal  level  for  the  new  load,  but  the  effect  is  not  as  serious 
as  if  the  surge  chamber  were  absent. 


152  HYDRAULICS 

In  Fig.  195  is  shown  a  surge  chamber  of  large  size.  It  is 
at  the  end  of  a  pressure  tunnel  which  is  approximately  7.76 
miles  in  length,  with  an  average  cross-section  area  of  100  sq. 
ft.  and  in  which  is  a  maximum  velocity  of  flow  of  10  ft.  per  sec.1 

EXAMPLES 

1.  A  cast-iron  pipe  line  is  24  in.  in  diameter  and  the  metal  is  0.75  in. 
thick.     If  the  velocity  of  water  in  it  is  6  ft.  per  sec.,  find  the  pressure  that 
would  be  created  by  the  instantaneous  closure  of  a  valve. 

Ans.  296.5  Ib.  per  sq.  in. 

2.  If  the  pipe  line  in  problem  (1)  were  500  ft.  long,  within  what  length 
of  time  must  the  valve  be  closed  to  produce  the  same  pressure  as  an  in- 
stantaneous closure?     What  would  the  length  of  time  be  if  it  were  .5,000 
ft.  long? 

Ans.  T  =  0.27  sec.,  2.7  sec. 

3.  If  the  pipe  line  in  problem  (1)  were  7,000  ft.  long  what  would  be  the 
time  of  closing  the  valve  so  that  the  pressure  produced  were  only  one- 
third  of  that  in  case  of  instantaneous  closure? 

Ans.  11.43  sec. 

108.  Relation  between  Absolute  and  Relative  Velocities. — In 

much  of  the  work  that  follows  it  will  be  necessary  to  deal  with 


FIG.  142. — Relation  between  absolute  and  relative  velocities. 

both  absolute  and  relative  velocities  of  the  water.  The  absolute 
velocity  of  a  body  is  its  velocity  relative  to  the  earth.  The  rela- 
tive velocity  of  a  body  is  its  velocity  relative  to  some  second 
body  which  may  in  turn  be  in  motion  relative  to  the  earth.  The 
absolute  velocity  of  the  first  body  is  the  vector  sum  of  its  velocity 
relative  to  the  second  body  and  the  absolute  velocity  of  the  lat- 
ter. The  relation  between  the  three  is  shown  in  Fig.  142.2 

1 W.    F.    Durand,    "Control   of   Surges  in   Water   Conduits,"   Journal, 
A.  S.  M.  E.,  June,  1911. 

See  also,  "The  Differential  Surge  Tank" 
by  R.  D.  Johnson,  Trans.  A.  S.  C.  E.,  vol.  78, 
page  760  (1915). 

FIG.  143.  2  A  clearer  idea  of  this  relationship  may  be  ob- 

tained from  the  illustration  in  Fig.  143.     Suppose 
a  raft  is  moving  downstream  with  a  uniform  velocity  u.     A  man  on  the 


HYDRODYNAMICS  153 

109.  Impulse  of  a  Jet. — When  a  stream  of  water  strikes  any 
object,  the  dynamic  force  exerted,  due  to  the  impact,  is  often 
termed  the  impulse  of  the  jet.  The  value  of  this  force  for  any 
case  can  be  determined  by  a  direct  application  of  equation  (83), 
or  some  of  its  special  forms. 

As  an  illustration  let  us  take  a  special  case,  shown  in  Fig.  144, 
where  the  jet  strikes  an  object  moving  with  a  uniform  veloc- 
ity in  the  same  direction  as  the 
jet.  Let  us  assume  the  x  axis 
parallel  to  the  jet  and  use  equa- 
tions (84)  and  (8.5).  We  shall 
assume  the  velocity  of  the  jet 
Vi,  its  cross-section  area  F},  the 

angle  a2,  and  the  velocity  of  the  object  u  to  be  known  quantities 
and  it  will  be  necessary  to  express  the  final  result  in  terms  of 
these. 

The  rate  of  discharge  of  the  jet  will  be  F\  V\  so  that  W  =  wFi  V\. 
But  this  may  not  be  the  amount  of  water  that  strikes  the  object 
per  unit  time.  If  the  body  is  moving  as  rapidly  as  the  jet  it  is 
seen  that  none  of  the  water  will  strike  it.  In  fact  the  measure 
of  the  amount  of  water  which  strikes  any  moving  object  is  the 
relative  velocity  Vi  for  this  is  the  velocity  with  which  the  water 
really  flows  upon  the  surface  of  the  object.  If  we  denote  by 
W  the  weight  of  water  striking  the  moving  body  per  second 
then  W  =  wFiV\.  In  this  special  case,  since  u  and  V\  are  in 
the  same  direction  we  have  v\  =  Vi  —  u.  But  in  general  we 
should  have  a  vector  relationship  here  as  shown  in  Fig.  142. 
However,  for  this  special  case,  W  =  wF\(V\  —  u).  The  reason 
that  less  water  strikes  the  body  per  second  than  comes  out  of  the 
orifice  or  nozzle  per  second  is  that  the  body  is  moving  away  from 
the  orifice  and  there  is  an  increasing  volume  of  water  between 
the  two. 

It  has  already  been  shown  that,  for  the  special  case  we  are 
considering,  v\  —  V\  —  u.  Now  in  flow  over  the  vane  the  water 
will  suffer  some  frictional  resistance  which  will  reduce  its  velocity, 

raft  at  A  walks  over  to  the  diagonally  opposite  corner  at  a  uniform  rate. 
But  by  the  time  he  reaches  B  the  latter  point  on  the  raft  will  have 
moved  downstream  to  point  C.  Thus  the  path  of  the  man  relative  to 
the  raft  is  AB  but  relative  to  the  earth  it  is  AC.  Since  the  velocities  are 
all  uniform  they  are  all  proportional  to  the  distances  traversed  in  this  in- 
terval of  time. 


154  HYDRAULICS 

so  that  vz  is  always  less  than  v\.  We  may  thus  say  that  vz  =  nvi, 
where  n  is  less  than  unity.  But  from  Fig.  142  it  may  be  seen 
that  F2  cos  A<z  =  u  -f  ^2  cos  a2. 

Replacing  v2  in  terms  of  v\  and  that  in  turn  in  terms  of  V\  and 
HI  we  have 

F2  cos  AZ  =  u  +  w(Vi  —  w)  cos  a2. 

Substituting  these  values  in  equation  (84)  we  have 

W 

Px  =  — (1  -  ncosa2)(7]  -  u)  (95) 

t7 

or 

p,  =  H^!  (1  -  n  cos  a,)(7i  -  w)2  (96) 

The  value  of  Py  may  be  determined  in  a  similar  manner.  It 
must  be  borne  in  mind  that  equations  (95)  and  (96)  are  true  only 
for  A1  =  0°. 

It  may  be  seen  that  the  magnitude  of  the  force  exerted  by  a 
jet  depends  upon  the  shape  and  velocity  of  the  object  struck. 

EXAMPLES 

1.  A  jet  of  water  3  in.  in  diameter  has  a  velocity  of  120  ft.  per  sec.     It 
strikes  a  vane  with  an  angle  a2  =  90°  which  moves  in  the  same  direction  as 
the  jet  with  a  velocity  u.     Assume  that  the  loss  in  flow  over  this  vane  is 
such  that  n  =  0.9.     When  u  has  values  of  0,  40,  60,  80,  100,  and  120  ft. 
per  sec.,  find  values  of:  (a)  W,  (6)  F2  cos  A2,  (c)  Px- 

2.  If  the  jet  in  the  preceding  problem  strikes  a  vane  for  which  a2  =  180°, 
all  other  data  remaining  the  same,  find  values  of:  (a)  W,  (6)  vz,  (c)  72, 

C<*)P. 

110.  Reaction  of  a  Jet. — When  a  stream  of  water  issues  from 
any  device,  such  as  the  vessel  shown  in  Fig.  145,  a  force  is  re- 
quired to  accelerate  the  water  and  impart  to  it  the  velocity 
it  has  upon  leaving.  This  force  is  exerted  upon  the  parti- 
cles of  water  flowing  out  the  orifice  by  adjacent  particles  of  water 
and  ultimately  by  the  walls  of  the  vessel.  By  the  law  of  action 
and  reaction  an  equal  and  opposite  force  will  be  exerted  upon 
the  vessel.  It  is  impossible  to  analyze  this  reaction  in  detail  but 
we  know  that  its  total  value  will  be  given  by  an  application  of 
equation  (83). 

Let  us  assume  that  the  vessel  in  Fig.  145  moves  to  the  left 
with  a  uniform  velocity  u,  and  that  the  orifice  is  so  small  com- 
pared to  the  size  of  the  vessel  that  the  relative  velocity  of  the 


HYDRODYNAMICS 


155 


water  in  the  latter  may  be  neglected  as  may  also  the  change  in  h. 
Then  V\  =  u.  If  the  jet  issues  from  the  orifice  with  a  velocity 
vz  the  absolute  velocity  of  the  jet  will  be  F2  =  u  —  vz.  Hence 
AF  =  Vi  —  V*  =  u  —  (u  —  v2)  =  vz.  Therefore 


wF 


(97) 


This  might  have  been  determined  more  directly  if  another 
proposition  had  been  previously  established.  That  is  that  for 
any  case  whatever  AFX  =  Avj.  +  Awx,  where  the  subscript  x 
merely  denotes  a  component  along  any  axis.  In  this  case,  since 
u  is  constant,  it  may  be  seen  that  AF 
=  At;  and  is  independent  of  the  velocity 
of  the  vessel. 

Since  vz  =  cv\/2gh,  we  may  write 
equation  (97)  as 


P  =  2cv*wFh 


(98) 


FIG.  145. 


If  losses  of  energy  be  neglected  in 
both  cases,  it  may  be  seen  that  the  re- 
action of  the  jet  in  Fig.  145  is  equal  to  the  force  of  impact  upon 
a  flat  plate,  normal  to  the  jet,  providing  the  area  of  the  plate  be 
large  enough  to  deflect  the  water  through  90°. l 

111.  Distinction  between  an  Impulse  and  a  Reaction  Turbine. 
—The  distinction  between  these  two  fundamental  types  of  tur- 
bines may  be  made  upon  the  basis  of  the  action  of  the  water  as 
defined  in  the  two  preceding  articles.  The  impulse  turbine  is 
acted  upon  by  the  impact  or  the  impulse  of  a  stream  of  water 
which  strikes  it.  The  working  force  in  the  case  of  the  reaction 
turbine  is  largely  that  due  to  the  reaction  of  the  streams  of  water 
as  they  are  discharged  from  the  runner. 

We  may  also  state  other  conditions  which  accompany  those 
just  given.  Thus  the  water  within  the  impulse  turbine  is  not 
confined  but  is  open  to  the  atmosphere,  while  in  the  reaction 
turbine  the  wheel  passages  must  be  completely  filled  with  water 
under  pressure.  The  energy  delivered  to  the  impulse  turbine  is 

1  The  hydrostatic  pressure  on  an  area  equal  to  that  of  the  jet  F  at  a  depth 
h  is  given  by  wFh.  The  fact  that  this  is  only  half  the  dynamic  pressure 
considered  is  of  no  significance.  As  has  already  been  pointed  out,  the 
dynamic  pressure  on  a  plate  is  distributed  over  an  area  much  larger  than 
that  of  the  jet  and  we  have  not  increased  the  intensity  of  pressure  in  either 
case. 


156 


HYDRAULICS 


all  kinetic,  while  that  delivered  to  the  reaction  turbine  is  partly 
kinetic  and  partly  "pressure  energy." 

But  it  is  well  to  bear  in  mind  that  in  both  types  the  essential 
thing  is  that  the  velocity  of  the  water  must  be  altered  in  order 
that  a  dynamic  force  may  be  exerted  upon  the  wheel.  And  in 
both  types  it  is  necessary  if  high  efficiency  is  attained  that  the 
absolute  velocity  of  the  water  as  it  leaves  the  wheel  be  low, 
since  this  velocity  represents  so  much  kinetic  energy  that  is  not 
utilized. 

112.  Theorem  of  Angular  Momentum. — In  Fig.  146  we  will 
suppose  a  particle  of  mass  dm  to  be  located  at  a  point  whose  co- 
ordinates are  x  and  y  and  to  be  moving  with  a  velocity  V.  The 
momentum  of  this  particle  will  be  dm.V.  The  moment  of  mo- 
mentum is  called  angular  momentum.  For  this  particle  the 
angular  momentum  is  dm.V  X  rcosA.  Since  the  moment  of 


-.'--r"'a 


X         O 

FIG.  146. 


any  quantity  is  the  algebraic  sum  of  the  moments  of  its  compo- 
nents, we  may  write 

dm.rV  cos  A  =  dm.VyX  —  dm.Vxy  =  dml-rrx  —  -rr  y  )  . 

Differentiating  the  above  with  respect  to  time  we  obtain 
d(rVcosA)        7    /dy  dx   .      dzy       dx  dy 


=  dm(ayx  —  axy) 

where  a  denotes  acceleration,  with  ax  and  ay  as  its  axial  com- 
ponents. (Vx  =  dx/dt,  ax  =  dVx/dt  =  d2x/dt2,  etc.).  If  the  re- 
sultant force  acting  on  the  particle  be  denoted  by  dR,  dmav  = 
dRV)  and  dmax  =  dRx.  Thus 

dmd(rV  cos  A)/dt  =  dRvx  -  dRxy. 


HYDRODYNAMICS 


157 


The  torque  exerted  upon  the  particle  with  respect  to  point  0 
is  seen  to  be  dR  X  L     By  the  principle  of  moments 


dR  X  I  = 


-  dRxy 


Thus,  if  T  denotes  torque  so  that  dR  X  I  =  dT, 
dT  =  dmd(rV  cos  A)/dt 


(99) 


That  is,  the  time  rate  of  change  of  the  angular  momentum  of  any 
particle  with  respect  to  an  axis  is  equal  to  the  torque  of  the  re- 
sultant force  on  the  particle  with  respect  to  the  same  axis. 

113.  Torque  Exerted  upon  Turbine  by  Water. — When  a  stream 
flows  through  a  turbine  runner  in  such  a  way  that  its  distance 
from  the  axis  of  rotation  remains  unchanged,  the  dynamic  force 
can  be  computed  by  the  principles  of  Art.  104.  But  when  the 


Eelartive  Path 
of  Water 


Absolute  Path 
of  Water 


FIG.  147. — Hydraulic  turbine. 

radius  to  the  stream  varies  it  is  not  feasible  to  compute  a  single 
resultant  force.  It  is  necessary  to  find  the  total  torque  exerted 
by  summing  up  the  elementary  torques  produced  by  all  the  ele- 
mentary forces. 

In  Fig.  147  let  MN  represent  a  vane  of  a  wheel  which  may 
rotate  about  an  axis  0  perpendicular  to  the  plane  of  the  paper. 
Water  enters  the  wheel  at  M  and  since  the  wheel  is  in  motion, 
by  the  time  the  water  arrives  at  N  on  the  vane  that  point  of  the 
vane  will  have  reached  position  N'.  Thus  the  absolute  path  of 
the  water  is  really  MN'. 

Let  us  consider  an  elementary  volume  of  water  forming  a  hol- 
low cylinder,  or  a  portion  thereof,  concentric  with  0.  Let  the 


158  HYDRAULICS 

area  of  this  normal  to  the  radius  be  Fr,  and  its  thickness  be  dr. 
Then  the  element  of  mass  will  be  dm  =  wFrdr/g.  Substituting 
this  value  in  equation  (99)  we  have 

,_       wFrdr  d(rV  cos  A)       wFrdr  ,,  T 
dT  =  -  :  —  rr~    -  =  —  TT-  d(rV  cos  A) 

g  dt  gdt 


But  dr/d£  =  Vr,  the  radial  component  of  velocity,  and  multiply- 
ing by  Fr,  which  is  the  cross-section  area  normal  to  r,  we  obtain  q, 
which  is  constant  for  steady  flow.  Thus  we  may  write 

W  f2 
T  =  —  J    d(rV  cos  A) 

Integrating  between  limits  we  have  the  value  of  the  torque  ex- 
erted by  the  wheel  upon  the  water,  or  by  changing  signs,  the 
value  of  the  torque  exerted  by  the  water  upon  the  wheel.  There- 
fore the  torque  exerted  upon  the  wheel  by  the  water  is 

W 

T  =  —  (nVi  cos  Ai  -  r2F2  cos  A2)  (100) 

y 

It  may  be  seen  that  V  cos  A  is  the  tangential  component  of 
velocity.  It  is  convenient  to  represent  this  by  a  single  letter  s 
and  so 

W 

T  =  ~  (rlSl  -  r2s2)  (101) 

*/ 

It  is  immaterial  in  the  application  of  this  formula  whether  the 
water  flow  radially  inward,  as  in  Fig.  147,  radially  outward,  or 
remains  at  a  constant  distance  from  O.  In  any  case  TI  is  the 
radius  at  entrance  and  r2  is  that  at  exit. 

114.  Torque  Exerted  upon  Water  by  Centrifugal  Pump.  —  The 
derivation  of  an  expression  for  the  torque  exerted  by  a  pump 
impeller  is  exactly  the  same  as  in  the  preceding  article  except 
for  the  substitution  of  limits.  As  turbines  are  universally  con- 
structed there  are  certain  guide  vanes  surrounding  the  runner 
which  give  the  water  its  initial  direction  AI.  Thus  any  angular 
momentum  which  the  water  has  as  it  flows  into  the  runner  is 
imparted  by  the  guides.  In  some  centrifugal  pumps  there  are 
guide  vanes  within  the  "eye"  of  the  impeller,  which  give  the 
water  a  definite  direction  as  it  flows  into  the  latter.  For  such 
pumps  we  should  merely  reverse  equation  (101),  since  we  desire 
the  torque  exerted  upon  the  water  and  not  by  it. 

But  for  the  usual  type  of  centrifugal  pump  there  is  nothing  at 


HYDRODYNAMICS  159 

entrance  to  the  impeller  to  give  the  water  any  definite  direction. 
In  fact  the  water  enters  a  given  impeller  at  different  angles  de- 
pending upon  conditions  of  operation,  and,  while  the  ordinary 
pump  is  designed  for  a  " radial  entrance"  this  can  be  had  only 
for  the  normal  rate  of  discharge.  But  any  angular  velocity  with 
which  the  water  enters  the  impeller  has  been  really  derived  from 
the  impeller  and  so  should  be  credited  to  the  latter.  Hence  we 
should  take  as  our  lower  limit  of  integration  not  (1)  where  the 
water  enters  the  impeller  but  some  point  back  in  the  suction  pipe, 
where  the  angular  momentum  is  zero.1  Thus  we  should  have 


Absolut 
Path. 


FIG.  148.  —  Centrifugal  pump. 

for  the  ordinary  centrifugal  pump  under  all  conditions  of  opera- 
tion, . 

W 

T  =  —  r2s2  (102) 

115.  Power.  —  If  torque  be  multiplied  by  angular  velocity 
the  product  represents  power.  Angular  velocity  must  be 
expressed  in  radians  per  second,  hence,  if  N  =  r.  p.  m.;  the  horse- 
power will  be 


If  T  has  the  value  given  by  equation  (101),  the  power  will  be 
less  than  that  supplied  to  the  turbine  by  the  water,  the  difference 
being  the  power  lost  in  hydraulic  friction  within  the  turbine  case, 
runner,  and  draft  tube.  It  is  greater  than  the  power  delivered 
by  the  turbine  by  an  amount  equal  to  the  losses  in  mechanical 
friction.  It  is  the  power  that  is  actually  delivered  to  the  shaft 

1  This  point  is  fully  discussed  by  the  author  in  "Centrifugal  Pumps," 
page  61. 


160  HYDRAULICS 

from  the  water,  and  is  analogous  to  the  indicated  power  of  a 
steam  engine. 

If  T  has  the  value  given  by  equation  (102)  the  power  given  by 
equation  (103)  will  be  less  than  that  required  to  run  the  pump 
by  an  amount  equal  to  the  mechanical  losses  and  it  will  be  greater 
than  the  power  delivered  in  the  water  by  the  amount  of  the  hy- 
draulic losses.  It  represents  the  power  actually  expended  by  the 
impeller  on  the  water  and  is  analogous  to  the  indicated  power  of 
a  reciprocating  pump. 

While  equations  (101)  and  (102)  are  true,  they  are  of  little  real 
service  because  the  proper  values  to  use  in  them  are  often  not 
known  with  exactness.  The  precise  values  of  velocities  and  direc- 
tions of  stream  lines  are  difficult  matters  to  determine.  Since 
water  does  not  fulfill  the  ideal  conditions  assumed,  it  will  be 
found  that  these  equations  often  yield  numerical  results  that  are 
considerably  in  error.1 

It  should  be  noted  that  power  can  be  expressed  in  the  following 
forms;  as  well  as  by  equation  (103). 

Pi/ 
Hp.  =  (104) 


WH       wqti         qH 
-  550  '  ^    - 


In  the  last  expression  H  may  represent  any  head  for  which  the 
corresponding  power  is  desired. 

116.  Definitions  of  Heads.  —  In  turbine  and  pump  practice  we 
find  the  word  "head"  used  to  express  several  different  physical 
quantities.      The  head  h  under  which  a  turbine  or  pump  actually 
operates  is  explained  in  Arts.  88  and  89.     But  there  is  energy 
lost  in  hydraulic  friction  within  the  runner  or  impeller  and  thus 
there  is  head  lost.      We  shall  designate  this  by  h'.     And  in  the 
turbine  a  portion  of  the  energy  of  the  water  is  delivered  in  the 
form  of  mechanical  work  and  the  head  thus  utilized  by  the  runner 
we  shall  denote  by  h"  '.     In  the  centrifugal  pump  h"  will  represent 
the  head  actually  imparted  to  the  water  by  the  impeller. 

Thus  for  the  turbine  we  shall  have  h"  =  h  —  h'  and  for  the 
pump  h  =  h"    -  h'. 

117.  Definitions   of  Turbine   Efficiencies.—  The   word    "  effi- 
ciency" without  any  qualifying  adjective  is  always  understood 

!See  "  Hydraulic  Turbines,"  page  83  and  "Centrifugal  Pumps,"  pages 
76,  81,  82,  and  84. 


H  YDROD  YNAMICS  161 

to  mean  gross  or  total  efficiency.  It  is  the  ratio  of  the  developed 
or  brake  horsepower  to  the  power  delivered  in  the  water  to  the 
turbine.  That  is 

e  =  b.hp./w.hp.  (106) 

Mechanical  efficiency  is  the  ratio  between  the  power  delivered 
by  the  machine  and  the  power  delivered  to  its  shaft  by  the  water. 
If  q  represents  the  total  turbine  discharge  while  q'  equals  the 
amount  of  leakage  through  the  clearance  spaces,  the  actual 
amount  of  water  doing  work  is  q  —  q'  .  Hence 

(107) 

Hydraulic  efficiency  is  the  ratio  of  the  power  actually  delivered 
to  the  shaft  to  that  supplied  in  the  water.  That  is 

eh  =  wqh"/wqh  =  h"/h  (108) 

Volumetric  efficiency  is  the  ratio  of  the  water  actually  used  by 
the  runner  to  total  amount  discharged.  Thus, 

e,  =  (q  ~  q')/q  (109) 

The  total  efficiency  is  the  product  of  these  three  separate  fac- 
tors. ^That  is, 

e  =  em  X  eh  X  ev  (110) 

118.  Definitions  of  Pump  Efficiencies.  —  The  various  pump 
efficiencies  are  similar  to  those  for  the  turbine.  'The  total  effi- 
ciency is 

e  =  w.hp  /b.hp.  (Ill) 

The  mechanical  efficiency  is 


The  hydraulic  efficiency  is 

eh  =  wqh/wqh"  =  h/h"  (113) 

The  volumetric  efficiency  is 

e,  =  q/(q  +  ?')  (114) 

As  in  equation  (110)  the  total  efficiency  is  the  product  of  these 
three. 

119.  Centrifugal  Action  or  Forced  Vortex.  —  If  a  vessel  con- 
taining a  liquid  is  rotated  about  'its  axis,  the  liquid  will  tend  to 
11 


162  HYDRAULICS 

rotate  at  the  same  speed.  If  we  take  an  elementary  volume  in 
Fig.  149  whose  length  along  the  radius  is  dr  and  whose  area 
normal  to  the  radius  is  dF,  we  have  an  elementary  mass  wdFdr/g 
moving  in  a  circular  path.  This  mass  has  an  acceleration 
uz/r  or  o>2r,  directed  toward  the  axis  of  rotation.  Consequently 
the  accelerating  or  resultant  force  is  (wdFdr/g)  co2r  directed  toward 
the  axis.  The  intensity  of  pressure  on  the  two  faces  of  the  ele- 
mentary volume  differs,  by  dp'  =  wdpr.  The  value  of  the 
resultant  force  is  therefore  wdprdF.  Consequently, 

wdprdF  = 
dpr  = 

But  this  expression  shows  only  the  difference  of  pressure  along 

the  radius  and  in  the  same  horizontal 
plane.  If  we  move  along  a  path  parallel 
to  the  vertical  axis  of  rotation  so  that 
the  radius  is  constant,  the  pressure  de- 
creases directly  as  the  elevation  in- 
creases. Thus, 

dpz  =  -  dz. 

The  variation  of  the  intensity  of  pres- 
sure in  any  direction  may  be  found  'by 
combining  the  two  preceding  equations. 

Thus,  in  general,  when  both  r  and  z 
FIG.  149. — Forced  vortex. 

vary, 

dp  =  -  dz  +  (co2/</)rdr  (115) 

To  find  the  equation  of  the  free  surface  or  any  surface  of  equal 
pressure  we  need  only  place  dp  equal  to  zero.     We  then  have 

fdz  =  (u*/g}frdr 

z  =  rza)2/2g  +  constant. 

To  determine  the  constant  we  may  assume  z  =  0  when  r  =  0. 
Thus  the  constant  =  0  so  that 

z  =  r2o>2/20  (116) 

From  this  it  may  be  seen  that  the  free  surface  or  any  surface  of 
equal  pressure  is  a  paraboloid. 

To  find  the  variation  of  pressure  in  the  same  horizontal  plane 


HYDRODYNAMICS 


163 


we  need  only  assume  dz  =  0,  and  integrating  between  limits  we 
obtain 

P2  -  Pi  =  (r22  -  ri2)«2/20  =  (uf  -  u^/2g         (117) 


For  the  difference  in  pressure  between  any  two  points  we  must 
integrate  equation  (115)  which  gives 

Pz  -  Pi  =  Zi  -  Z2  +  u22/2g  -  Ul2/2g  (118) 

120.  Free  Vortex.  —  Where  external  forces  are  applied  to  the 
water,  as  in  the  preceding  case,  we  have  a  forced  vortex.  Where 
no  external  forces  are  applied  but  the  water  rotates  by  virtue  of 
its  own  angular  momentum  previously  derived,  we  have  a  free 
vortex.  Let  us  first  consider  a  pure  radial  flow  between  two 
parallel  plates  as  in  Fig.  150.  .  In  such  a  case  V  =  Vr,  where  Vr 
denotes  the  radial  component  of  the  velocity.  If  b  is  the  constant 
distance  between  the  two  plates  we  have 

q  =  2<jrrbVr 

from  which  it  may  be  seen  that  Vr  varies  as  1/r. 

A  free  circular  vortex  consists  of  a  body  of  water  in  rotation 
without  any  appreciable  flow. 
Thus  the  stream  lines  are  con- 
centric circles  and  V  —  s. 
Since  no  torque  is  exerted 
on  the  water,  it  follows  that 
there  can  be  no  change  in  an- 
gular momentum.  Since  an- 
gular momentum  is  propor- 
tional to  s  X  r,  it  is  apparent 
that  s  varies  as  1/r,  if  the 
angular  momentum  is  constant. 

A  free  spiral  vortex  is  a  combination  of  radial  flow  and  circular 
flow.  Since  V2  =  Vr2  +  s2,  it  may  be  seen  that  for  spiral  stream 
lines  V  also  varies  as  1/r.  Since  7,  Vr,  and  s  all  vary  as  1/r,  it 
may  be  seen  that  the  angle*  between  V  and  s  remains  constant. 
Thus  the  stream  line  must  be  a  logarithmic  spiral. 

Since  no  energy  is  imparted  to  the  water,  we  will  have,  if  fric- 
tion is  neglected,  H  —  p  +  2  +  V2/2g  =  constant,  or 

H  =  pi  +  VS/2g  =  pz  +  V2*/2g 

P2  =  Pi  +  [1  -  (ri/r2)WA7  (119) 

As  r  increases  it  may  be  seen  that  p  approaches  H  as  a  limit. 
The  last  portion  of  equation  (119)  is  the  maximum  possible  gain 


FIG.  150.— Free  vortex. 


164  HYDRAULICS 

of  pressure  in  a  free  spiral  vortex.  Actually,  owing  to  frictional 
losses  and  instability  of  flow  resulting  in  a  departure  from  the 
stream  lines  assumed,  the  pressure  increase  will  be  much  less. 

121.  PROBLEMS 

1.  Find  the  horsepower  of  a  jet  of  water  with  a  cross-section  area  of  3 
sq.  in.  if  it  has  a  velocity  of  100  ft.  per  sec. 

Ans.  36.8  hp. 

2.  Suppose  this  jet  in  problem  (1)  were  to  strike  a  wheel  with  curved 
vanes.     Assume  that  A\  =0°,  r\  =  r^,  and  that  the  vanes  reversed  the 
relative  velocity  of  the  water  through  180°  without  friction  loss.     Find 
values  of  the  force  exerted  when  the  peripheral  speeds  of  the  vanes  are 
0,  30,  50,  80,  and  100  ft.  per  sec.     (For  an  entire  wheel  we  use  W  and  not 
W.) 

Ans.  808,  566,  404,  161.5,  and  0  lb.,  respectively. 

3.  Find  the  horsepower  for  the  five  speeds  given  in  problem  (2). 
Ans.  0,  30.8,  36.8,  23.5,  and  0  hp.,  respectively. 

4.  What  are  the  efficiencies  of  the  wheel  in  problem  (2)  at  the  various 
speeds  given?     When  the  power  of  the  wheel  is  less  than  that  of  the  jet,  what 
becomes  of  the  difference? 

6.  Suppose  that  the  wheel  in  problem  (2)  were  equipped  with  vanes  for 
which  «2  =  90°.  How  would  the  values  of  force  and  power  compare  with 
the  values  when  a2  =  180°? 

6.  Suppose  that  the  wheel  in  problem  (2)  were  equipped  with  vanes  for 
which  «2  =  160°,  and  that  the  loss  in  flow  over  the  vanes  were  such  that 
vz  =0.8  VL     Find  the  values  of  force  exerted,  power,  and  efficiency  for  the 
five  speeds  given. 

7.  What  would  be  the  force  of  the  reaction  of  the  jet  in  problem  (1)? 

8.  The  absolute  velocity  of  water  entering  a  turbine  runner  is  60  ft.  per 
sec.  and  that  leaving  is  15ft.  per  sec.     Ai  =  20°,  Aa  =  80°,  n  =  2.5  ft., 
r2  =4.0  ft.     (a)  If  W  =  600  lb.  per  sec.,  find  the  torque  on  the  wheel. 
(6)  If  HI  =  50  ft.  per  sec.,  find  the  power  delivered  to  the  wheel. 

Ans.  (a)  2,430  ft.-lb.     (6)  88.5  hp. 

9.  If  the  radius  =  3  ft.,  find  the  torque  exerted  on  the  wheel  in  problem 
(2).     Find  it  by  using  equation  (101)  and  compare  with  values  obtained  by 
multiplying  P  and  r. 

10.  An  open  cylindrical  vessel  is  rotated  about  its  axis,  which  is  vertical. 
If  the  vessel  is  partially  filled  with  water,  what  speed  would  be  necessary 
to  cause  the  water  surface  at  a  radius.of  1.5  ft.  to  be  4.0  ft.  higher  than 
the  surface  at  the  center  of  rotation? 

11.  A  closed  vessel  completely  filled  with  water  is  rotated  about  its  axis 
at  a  speed  of  2,000  r.p.m.     If  the  pressure  at  the  center  of  rotation  is  2  ft. 
of  water,  what  will  it  be  at  a  radius  of  6  in.  ? 

Ans.  172  ft. 

12.  If  the  inner  and  outer  radii  of  the  "whirlpool  chamber"  in  Fig.  150 
are  8  in.  and  16  in.  respectively,  what  will  be  the  values  of  V,  Vr,  and  s  at 
the  outer  diameter  when  water  enters  the  inner  diameter  with  a  velocity  of 
80  ft.  per  sec.  at  an  angle  of  15°  with  the  tangent?     What  will  be  the  gain 
of  pressure,  neglecting  losses? 


CHAPTER  X 
DESCRIPTION  OF  THE  IMPULSE  WHEEL 

122.  The  Impulse  Wheel.— There  have  been  several  types  of 
impulse  turbines  produced,  but  the  only  one  that  has  survived 
in  this  country  is  of  the  kind  shown  in  Fig.  151.  This  is  the 
impulse  wheel  or  the  Pelton  wheel,  so  called  in  honor  of  L.  A. 
Pelton  who  contributed  to  its  early  development.  It  may  be  also 
designated  by  the  name  of  the  tangential  waterwheel,  from  the 
fact  that  the  center  line  of  the  jet  is  tangent  to  the  path  of  the 
center  of  the  buckets. 


From  a  photograph  by  the  author. 

FIG.  151. — Impulse  water  wheel  with  needle  nozzle, 
and  nozzle  is  wide  open.) 


(Needle  is  drawn  back 


The  wheel  in  Fig.  151  is  operated  by  a  jet  of  water  from  the 
nozzle  at  the  left.  This  same  wheel  in  action  may  be  seen  in 
Figs.  206  to  210.  A  view  of  another  wheel  showing  the  relation 
of  the  nozzle  to  the  buckets  is  shown  in  Fig.  152.  The  jet  strikes 
the  dividing  ridge,  or  "  splitter,"  of  the  buckets,  is  divided  into 
two  parts,  flows  over  the  face  of  the  bucket,  and  is  finally  dis- 
charged at  both  sides  of  the  latter. 

165 


166 


HYDRAULICS 


In  Figs.  153  and  154  we  see  views  of  an  assembled  wheel  with 
the  " chain  type"  of  construction.  That  is,  each  bolt  is  instru- 
mental in  holding  two  buckets,  so  that  the  latter  are  fastened 
together  as  a  chain.  This  permits  of  a  compact  construction 
and  enables  the  buckets  to  be  placed  closer  together  than  in 
the  type  shown  in  Fig.  151. 


From  a  photograph  by  the  author. 

FIG.  152. — Impulse  wheel  viewed  from  below.     (Nozzle  closed  by  needle.) 
D  =  84";  h  =  134';  N  =  124;  Hp  =  280. 


The  device  shown  at  the  right  in  Fig.  151  is  the  "stripper," 
its  function  being  to  prevent  water  being  carried  around  with 
the  wheel  and  thus  adding  to  the  windage  losses.  The  buckets 
pass  through  an  opening  in  this  with  a  clearance  of  about  0.5  in. 

123.  Buckets. — Typical  styles  of  buckets  now  in  use  for 
impulse  wheels  are  seen  in  Figs.  155  and  156.  The  theory 
shows  that  the  face  of  the  bucket  should  be  a  surface  of  double 


DESCRIPTION  OF  THE  IMPULSE  WHEEL       167 

curvature,  and  it  is  also  found  that  the  shape  of  the  back  of  the 
bucket  may  be  as  important  as  that  of  the  face.  The  reason  for 
this  is  that  the  back  of  the  bucket  may  interfere  with  the  water 
which  is  acting  upon  the  bucket  ahead,  for  when  a  bucket 
swings  down  into  the  jet  it  merely  cuts  off  the  jet  from  the  preced- 
ing bucket  and  leaves  a  "slug"  of  water  to  complete  its  work  on 
the  one  ahead.  If  the  back  of  the  bucket  is  not  properly  shaped 
it  may  not  leave  sufficient  clearance  for  the  water.  The  "  notch  " 


From  a  photograph  by  the  author, 

FIG.   153.— Pelton-Doble  wheel,  in  shop  of  Pelton  Water  Wheel  Co. 
76";  h  =  540';  N  =  257;  Hp  =  2100. 


D 


is  cut  out  of  the  Pelton-Doble  bucket  so  that  it  may  reach  a  posi- 
tion where  its  path  is  more  nearly  tangent  to  that  of  the  jet 
before  the  latter  strikes  it.1 

For  service  under  moderate  heads  these  buckets  may  be  made 

1  For  impulse  wheels  of  high  specific  speeds  there  are  other  reasons  for 
this  construction  which  space  forbids  taking  up  here  in  detail.  In  brief,  it 
is  so  that  every  bit  of  water  may  complete  its  work  upon  the  bucket  before 
the  latter  leaves  the  line  of  action  of  the  jet,  in  which  event  some  of  the 
water  would  not  be  utilized.  (See  Fig.  202.) 


168 


HYDRAULICS 


of  cast  iron,  though  the  better  ones  are  of  bronze  or  steel.  For 
very  high  heads  only  the  latter  may  be  employed.  The  working 
face  of  the  bucket  should  be  smoothed  up  or  polished  and  the 
dividing  edge,  or  "splitter,"  ground  to  a  knife  edge  in  order  to 
reduce  hydraulic  friction  losses. 

For  high  efficiency  it  is  desirable  that  the  bucket  reverse  the 
relative  velocity  of  the  jet  as  nearly  as  is  feasible.     But  a  com- 


From  a  photograph  by  the  author, 

FIG.  154. — Pelton-Doble  wheel. 

plete  reversal  of  180°  is  not  permissible,  as  the  water  must  be 
thrown  to  one  side  so  as  to  clear  the  following  bucket.  An  angle 
of  about  165°  is  usually  employed,  though  even  170°  may  fre- 
quently be  used.  Due  to  surface  tension  the  actual  direction  of 
the  water  will  always  be  somewhat  less,  than  the  bucket  angle, 
the  difference  between  the  two  decreasing  as  higher  heads  are 
used.  For  good  efficiency  the  width  of  the  bucket  should  be  at 
least  three  times  the  diameter  of  the  jet,  and  the  diameter  of  the 
wheel  should  be  at  least  nine  times  that  of  the  jet.  (The  usual 


DESCRIPTION  OF  THE  IMPULSE  WHEEL       169 


From  a  photograph  by  the  author. 

FIG.   155. — Pelton-Doble  ellipsoidal  bucket. 


Courtesy  Allis-Chalmers  Mfg.  Co. 

FIG.  156, — Allis-Chalmers  buckets. 


170 


HYDRAULICS 


ratio  is  12  in  the  latter  case.)  Since  jets  10  in.,  or  more  in 
diameter,  are  in  use,  buckets  of  at  least  30  in.  in  width  are 
sometimes  seen. 

124.  Nozzles  and  Governing. — The  jets  used  in  impulse  wheels 
are  almost  always  furnished  by  needle  nozzles,  of  which  the 
earliest  type  is  shown  in  Fig.  158.  The  needle  of  the  style  used 
today  is  shown  in  Fig.  159.  As  it  is  moved  back  and  forth  in 
the  nozzle  it  varies  the  size  of  the  nozzle  opening  and  hence 
varies  the  amount  of  water  discharged.  But  fortunately  it 
does  not  involve  any  serious  loss  of  head  until  the  nozzle  is 
nearly  closed.  The  efficiency  of  a  needle  nozzle  when  it  is  wide 


Courtesy  Allis-Chalmers  Mfg.  Co. 

FIG.  157. — Allis-Chalmers  impulse  wheel  for  Pacific  Light  &  Power  Co. 
D  =  94";  h  =  I860';  N  =  375;  Up  =  10,000. 

open  may  be  about  97  or  98  per  cent.,  the  velocity  coefficient 
being  about  0.99  or  a  little  less.  The  nozzle  efficiency  would  not 
fall  below  90  per  cent,  until  the  needle  was  closed  so  far  that 
about  half  the  maximum  amount  of  water  was  being  discharged. 
Thus  it  is  a  very  efficient  regulating  device. 

In  order  to  keep  the  speed  of  a  wheel  constant  under  different 
loads  it  is  necessary  to  vary  the  amount  of  water  so  that  the 
power  supplied  to  the  turbine  shall  be  proportional  to  the  power 


DESCRIPTION  OF  THE  IMPULSE  WHEEL       171 


From  a  photograph  by  the  author. 

FIG.  158. — The  original  needle  nozzle. 


Courtesy  of  Pelton  Water  Wheel  Co. 

FIG.  159. — Pelton-Doble  needle  and  nozzle  tip. 


172  HYDRAULICS 

demanded.  This  can  sometimes  be  done  by  changing  the  posi- 
tion of  the  needle  in  accordance  with  the  power  the  wheel  must 
deliver.  Under  certain  conditions  the  governor  may  control  the 
position  of  the  nee.dle  for  this  purpose.  But  if  the  changes  of 
load  are  rapid  and  the  pipe  line  is  long,  this  procedure  would 
involve  serious  water  hammer,  if  close  speed  regulation  were 
attempted. 

In  order  to  secure  close  speed  regulation  and  yet  be  free  from 
the  danger  of  water  hammer,  the  deflecting  nozzle  is  often  used. 
The  entire  nozzle  is  movable  about  a  ball  and  socket  joint  near 
the  base  and  swings  on  trunnions.  In  case  of  a  sudden  drop  of 


From  a  photograph  by  the  author. 
FIG.  160. — Deflecting  needle  nozzle  for  a  10,000  h.p.  jet. 

load  on  the  machine  the  governor  could  lower  the  end  of  the 
nozzle  so  that  only  a  small  part  of  the  jet  struck  the  buckets, 
the  rest  of  the  water  being  wasted.  As  the  load  increased  the 
nozzle  could  be  raised  so  that  a  larger  amount  of  water  would 
strike  the  wheel.  As  this  would  be  wasteful  of  water,  such  noz- 
zles are  almost  always  equipped  with  needles  as  well,  which  can  be 
set  by  the  station  attendant  in  accordance  with  the  load  the 
wheel  carries.  Thus  water  would  be  wasted  for  a  short  time 
only,  but  the  needle  would  be  closed  so  slowly  that  no  damage 
would  be  done  to  the  pipe  line.  But  the  nozzle  may  be  deflected 


DESCRIPTION  OF  THE  IMPULSE  WHEEL        173 

with  any  degree  of  rapidity  so  that  close  speed  regulation  may 
be  secured.  Of  course  in  case  of  an  increase  in  load  it  would  be 
necessary  for  the  operator  to  open  the  nozzle,  as  the  governor  is 
powerless  there.  ^But  the  experience  is  that  increases  of  load 
come  on  gradually  enough  for  this  to  be  done.  The  chief  func- 
tion of  the  governor  is  to  prevent  racing  in  cases  of  abrupt  de- 


From  a  photograph  by  the  author. 
FIG.  161. — The  needle  nozzle  with  auxiliary  relief. 

creases  in  load.     Occasionally  the  nozzle  is  so  made  that  the 
governor  deflects  it  first  and  then  slowly  closes  the  needle. 

The  needle  nozzle  with  an  auxiliary  relief,  as  shown  in  Fig.  161, 
is  frequently  used.  In  this  type  the  jet  from  the  upper  nozzle 
strikes  the  wheel  while  that  from  the  lower  nozzle  goes  below  it. 


174 


HYDRAULICS 


Courtesy  of  Allis-Chalmers  Mfg.  Co. 

FIG.  162. — Needle  nozzle  with  deflecting  tip. 


From  a  photograph  by  F.  H.  Fowler. 
FIG.  163. — DeSabla  power  plant  in  normal  operation.    Under  head  of  1531  ft. 


DESCRIPTION  OF  THE  IMPULSE  WHEEL       175 

It  is  so  arranged  that  when  the  governor  closes  the  upper  nozzle 
it  opens  the  lower  one.  Thus  there  is  no  abrupt  change  in  flow 
in  the  pipe  line  as  the  surplus  water  simply  flows  out  through 
another  place.  But  in  order  to  prevent  waste  of  water,  the  con- 
nection between  the  governor  and  the  auxiliary  nozzle  is  a  dash- 
pot  arrangement  which  permits  the  needle  to  be  moved  only 
when  the  governor  movement  is  rapid,  and  when  the  relief 
has  been  opened,  this  arrangement  permits  it  to  be  gradually 


From  a  photograph  by  F.  H.  Fowler. 

FIG.  164. — DeSabla  power  plant  with  nozzles  deflected. 

closed  again.  Thus  we  have  accomplished  close  speed  regulation 
and  have  also  secured  economy  in  the  use  of  water. 

The  nozzle  shown  in  Fig.  162  is  similar  in  principle  to  the  de- 
flecting nozzle  in  that  the  jet  is  deflected  below  the  wheel.  But 
it  is  so  constructed  that  only  the  tip  of  the  nozzle  has  to  be  moved 
rather  than  the  entire  nozzle.  This  has  certain  advantages. 

It  will  be  noted  that  all  of  these  devices  may  prevent  rapid 
changes  in  flow  in  the  pipe  line  in  the  case  of  decreasing  loads. 


176 


HYDRAULICS 


But  only  a  surge  chamber  located  near  the  wheels  will  be  able 
to  supply  water  in  the  case  of  a  sudden  demand. 

125.  Conditions  of  Service. — The  impulse  wheel  is  well 
adapted  for  service  under  high  heads,  though  it  may  also  be 
employed  under  low  heads  if  the  power  is  small.  In  fact  the 
choice  of  the  type  of  turbine  is  a  function  of  power  as  well  as 
head. 

The  highest  head  that  has  ever  been  developed  is  in  Switzer- 
land where  15,000  hp.  is  generated  under  a  head  of  5,412  ft. 
The  power  of  each  wheel  in  the  plant  is  3,000  hp.,  its  diameter 


Courtesy  of  Allis-Chalmers  Mfg.  Co. 

FIG.  165. — Double  overhung  Allis-Chalmers  wheels  for  Pacific  Light  and 
Power  Co.     D  =  94";  h  =  I860';  N  =  375;  Hp  =  20,000  (for  unit). 

is  11.5  ft.  and  it  runs  at  500  r.p.m.  The  diameter  of  the  jet  is 
1.5  in. 

In  this  country  the  highest  head  that  has  been  used  is  2,100  ft., 
and  heads  ranging  from  1,000  ft.  to  2,000  ft.  are  not  uncommon. 
A  plant  under  2,100  ft.  static  head  is  shown  in  Fig.  198. 

The  jets  used  upon  impulse  wheels  are  of  all  sizes  up  to  about 
10  in.  or  a  little  over.  Ordinarily  only  one  jet  is  used  with  a 
single  wheel,  but  occasionally  two  or  more  nozzles  may  be  em- 
ployed, though  at  a  slight  sacrifice  of  efficiency.  In  order  to 
increase  the  power  of  a  single  unit  two  separate  wheels  are 
often  used  on  the  same  shaft,  as  in  Fig.  165. 

The  largest  power  developed  by  a  single  impulse  wheel  with 
one  jet  upon  it  is  10,000  hp.  A  wheel  of  that  capacity  is  shown 
in  Fig.  157,  though  there  are  several  cases  where  the  power  of  a 
single  wheel  has  approached  such  a  value. 


CHAPTER  XI 
DESCRIPTION  OF  THE  REACTION  TURBINE 

126.  The  Reaction  Turbine. — The  inward-flow  type  of  reac- 
tion turbine  is  the  only  one  that  is  of  any  importance  at  the 
present  time,  all  others  having  been  eliminated  because  of  certain 
relative  disadvantages.  Modern  inward-flow  turbines  are  com- 
monly known  as  Francis  turbines  in  honor  of  James  B.  Francis, 
who  built  the  first  successful  one  in  1849.  However,  the  wheels 
of  today  differ  considerably  from  his,  which  was  a  purely  radial- 
flow  turbine. 


Courtesy  of  Platt  Iron  Wks.  Co. 

FIG.   166. — Turbine  with  cylinder  gates  for  open  flume. 

By  radial  flow  is  meant  that  a  particle  of  water,  during  its 
flow  through  the  rotating  runner,  remains  in  a  plane  which  is 
normal  to  the  axis  of  rotation,  so  that  its  position  changes  only 
with  respect  to  its  distance  from  the  axis  of  rotation.  In  the 
evolution  of  the  modern  turbine  it  became  desirable  to  have  the 
water  enter  the  runner  with  a  "radial"  flow  and  then  to  turn  and 
flow  in  such  a  manner  that  a  component  of  its  velocity  might  be 
parallel  to  the  shaft.  In  fact  some  of  the  particles  of  water,  at 
12  177 


178 


HYDRAULICS 


least,  before  they  reached  the  discharge  edge  of  the  bucket  or 
vane  might  be  following  paths  which  lay  on  the  surfaces  of  cylin- 
ders concentric  with  the  axis.  This  is  known  as  mixed  flow,  and 
such  a  type  of  turbine  is  sometimes  called  the  American  turbine, 
though  the  name  Francis  is  generally  extended  to  cover  all  inward- 
flow  wheels.  In  Figs.  167  and  168  may  be  seen  the  nearest 
approach  in  present  practice  to  a  radial-flow  runner,  while  in 
Figs.  171  and  172  may  be  seen  the  mixed-flow  type. 

The  general  arrangement  of  a  reaction  turbine  may  be  seen  in 
Fig.  166.     This  particular  one  is  of  the  open-flume  type  and  is 


Courtesy  of  Pelton  Water  Wheel  Co 

FIG.  167. — Low-speed    turbine   runner       D 

Hp  =  20,000 


74" 


487';    N  =  360; 


set  so  as  to  be  completely  surrounded  by  water  in  a  manner 
similar  to  the  vertical-shaft  turbine  shown  in  Fig.  181.  The 
water  flows  through  the  stationary  guide  vanes  and  enters  the 
runner,  which  is  in  the  center.  During  flow  through  the  runner, 
the  velocity  of  the  water  suffers  a  change  in  both  direction  and 
magnitude  and  thereby  exerts  a  dynamic  force.  In  Fig.  166 
there  are  two  runners  set  on  the  same  shaft  and  discharging  into 
a  common  draft  chest,  from  which  the  water  flows  down  to  the 
tail  race  through  a  draft  tube. 


DESCRIPTION  OF  THE  REACTION  TURBINE    179 

127.  Runners. — The  part  of  the  turbine  upon  which  the  water 
does  its  work  is  called  the  runner.  Runners  may  be  built  up 
of  separate  pieces  of  metal  which  are  welded  together  but  they 
are  usually  cast  in  one  piece.  Occasionally  they  are  built  in 
sections  and  the  sections  bolted  together.  For  large  sizes  and 
low  heads  cast  iron  is  employed.  Better  runners  are  made  of 
bronze  and  occasionally  cast  steel  is  used  for  high  heads. 


From  a  photograph  by  the  author. 

FIG.  168. — Low  speed  turbine  runner  for  Pacific  Gas  and  Electric  Co. 

Runners  differ  considerably  in  their  proportions  and  appear- 
ances. One  extreme  is  shown  in  Figs.  167  and  168  while  the 
other  extreme  is  shown  in  Figs.  170,  172,  and  173.  It  may  be 
noted  that  the  runners  in  Figs.  167  and  172  develop  the  same 
amount  of  power  though  differing  widely  in  size.  This  is  due  to 
the  fact  that  the  smaller  runner  operates  under  a  much  higher 
head  and  consequently  needs  to  discharge  less  water  for  the 
same  amount  of  power.  And  the  largest  runner  in  the  world, 


180 


HYDRAULICS 


FIG.   169. — Double  discharge  turbine. 


From  a  photograph  by  the  author, 

FIG.  170. — The  largest  turbine  runner  in  the  world.     For  Cedars  Rapids 
Mfg.  and  Power  Co.     D  =  143";  h  =  30';  N  =  55.6;  Hp  =  10,800. 


DESCRIPTION  OF  THE  REACTION  TURBINE    181 

shown  in  Fig.  170,  develops  less  power  than  either  of  the  others 
because  it  is  under  a  still  lower  head. 


Ring 


FIG.  171. 


Courtesy  of  I.  P.  Morris  Co. 

FIG.    172.— High-speed    turbine   runner.     D  =  102";   h  =  76';    N  =  120; 

Hp  =  20,000. 

It  may  be  noted  that  the  width  of  the  runner  parallel  to  the 
shaft  in  Fig.  168  is  a  very  much  smaller  proportion  of  the  diame- 
ter of  the  runner  than  in  the  type  shown  in  Figs.  170  and  173. 


182 


HYDRAULICS 


Sometimes  runners  are  of  the  double  discharge  type  as  in  Fig. 
169  which  is  equivalent  to  placing  two  single  discharge  runners 
back  to  back.  Such  a  turbine  must  have  two  separate  draft 
elbows. 

As  shown  in  Fig.  171  there  may  be  several  places  at  which  the 
diameter  of  a  turbine  runner  may  be  measured  and  practice 
differs  in  this  respect.  The  custom  that  is  generally  followed  is 
to  give  the  mean  diameter  at  entrance  to  the  runner.  This  is 
the  dimension  that  will  be  found  in  Figs.  170  and  172.  The 


Courtesy  of  I.  P.  Morris  Co. 

FIG.  173. — Turbine  runner  for  Laurentide  Co. 

maximum  diameters  in  these  two  cases  are  17  ft.  7  in.  and  12  ft. 
7  in.  respectively. 

128.  Gates  and  Governing. — The  quantity  of  water  passed 
through  the  turbine  is  regulated  by  means  of  gates,  of  which  there 
are  several  kinds.  In  Fig.  166  we  find  the  cylinder  gate  used. 
In  that  class  of  turbine  the  guide  vanes  surrounding  the  runner 
are  absolutely  fixed.  Between  the  ends  of  these  vanes  and  the 
runner  is  a  metal  cylinder  which  may  slide  along  parallel  to  the 
shaft.  If  moved  in  one  direction  it  admits  water  to  the  runner 
and  may  be  so  far  withdrawn  as  to  offer  no  obstruction  whatever 


DESCRIPTION  OF  THE  REACTION  TURBINE    183 

between  the  guides  and  the  wheel.  And  if  it  is  moved  in  the 
other  direction  it  is  possible  to  shut  off  the  water  altogether. 
This  style  of  regulation  causes  the  turbine  to  have  a  poor  effi- 
ciency on  "part  gate,"  which  is  the  term  used  when  the  turbine 
is  running  under  less  than  full  load.  But  such  a  style  of  gate 
permits  a  turbine  to  be  constructed  at  less  cost. 


Courtesy  of  Pelton  Water  Wheel  Co. 

FIG.  174. — Wicket  gates  or  swing  gates. 

The  better  type  of  gate  so  far  as  efficiency  is  concerned  is  the 
kind  shown  in  Fig.  174.  Here  the  guide  vanes  themselves  are 
movable  and  by  rotating  about  their  axes  they  may  vary  the 
size  of  the  area  through  which  water  may  flow.  This  means  that 
the  angle  AI  changes.  These  gates  are  known  either  as  swing 
gates,  wicket  gates,  or  pivoted  guide  vanes.  They  involve  more 
expensive  construction  than  the  cylinder  gate  but  are  vastly 
better  if  economy  of  water  is  any  object. 


184 


HYDRAULICS 


In  Fig.  175  may  also  be  seen  some  movable  gates  as  they  are 
installed  in  the  turbine.  The  runner  is  to  go  into  the  space  in 
the  center. 


Courtesy  of  Petton  Water  Wheel  Co. 

FIG.  175. — Spiral-case  turbine  showing  swing  gates. 


Courtesy  of  Platt  Iron  Wks.  Co. 

FIG.   176. — Shifting  ring  for  operating  gates. 

The  swing  gates  are  operated  by  moving  a  "shifting  ring"  to 
which  each  gate  is  attached  by  links.     In  Fig.  176  may  be  seen 


DESCRIPTION  OF  THE  REACTION  TURBINE    185 

the  rods  from  the  governor  connected  to  this  ring  so  that,  when 
it  is  moved  slightly  with  the  turbine  shaft  as  a  center  of  rotation, 
each  gate  will  be  turned  through  some  angle.  The  links  which 
connect  the  gates  to  this  ring  can  be  seen  more  clearly  in  Fig.  177. 
The  problem  of  governing  a  reaction  turbine  is  similar  to  that 
of  the  impulse  wheel.  When  the  governor  closes  the  gates  and 
thus  reduces  the  discharge  through  the  turbine  it  is  necessary 
to  provide  some  bypass  for  the  water  in  order  to  prevent  water 
hammer  in  the  pipe  line.  The  usual  practice  is  to  use  a  relief 


Courtesy  of  Allis-Chalmers  Mfg.  Co. 

FIG.  177. — Swing  gates. 

valve  such  as  that  shown  over  at  the  right  in  Fig.  178.  When 
the  governor  closes  the  gates  it  opens  the  relief  valve  at  the  same 
time,  and  the  water  coming  down  the  pipe  is  then  discharged 
through  this  into  the  tail  race  alongside  the  draft  tube.  The 
action  of  such  a  relief  valve  may  be  seen  in  Figs.  179  and  180. 

The  connection  between  the  governor  and  the  relief  valve  is 
usually  not  a  rigid  connection,  in  order  that  the  relief  valve  may 
slowly  close  and  prevent  the  waste  of  water. 

129.  The  Draft  Tube. — The  water  is  conducted  from  the  tur- 
bine to  the  tail  race  through  a  draft  tube,  which  may  be  con- 


186  HYDRAULICS 

structed  of  riveted  steel  plates  as  in  Fig.  179  or  may  be  molded  in 
concrete  as  in  Fig.  182.  The  draft  tube  should  be  made  air- 
tight so  that  a  partial  vacuum  can  exist  within  it  and  thus  there 
may  be  a  "suction"  produced  on  the  discharge  side  of  the  runner 
which  shall  compensate  for  the  elevation  of  the  latter  above  the 
tail  water  level.  By  the  use  of  the  draft  tube  it  is  possible  to  set 


Courtesy  of  S.  Morgan  Smith  Co. 

FIG.  178. — Tallulah  Falls  turbine  showing  gate  mechanism  and  relief  valve 
H  =  580';  N  =  514;  Hp  =  19,000. 

the  turbine  at  a  convenient  distance  above  the  water  level  with- 
out losing  any  head  thereby. 

But  this  is  not  the  sole  function  of  the  draft  tube.  The  velocity 
with  which  the  water  is  discharged  from  the  runner  represents 
kinetic  energy  that  is  not  utilized  and  such  a  loss  cuts  down  the 
efficiency  of  the  wheel.  If  the  draft  tube  is  made  to  diverge  the 
velocity  at  its  mouth  will  be  much  less  than  that  with  which  the 
water  enters  it  from  the  runner  and  hence  the  kinetic  energy 
finally  lost  may  be  much  reduced.  With  some  types  of  turbine 
runners  it  is  necessary  to  allow  the  water  to  be  discharged  with 


DESCRIPTION  OF  THE  REACTION  TURBINE    187 

a  relatively  high  velocity  and  such  wheels  would  not  possess 
favorable  efficiencies  if  it  were  not  for  the  use  of  suitable  draft 
tubes.  The  usual  rate  of  diffusion  provided  for  is  such  that  a 
circular  tube  will  be  made  a  frustum  of  a  cone  the  vertex  angle 
of  which  is  8°.  Some  experiments  by  the  author  indicate  that 
a  larger  angle  than  this  might  be  permissible.  For  a  given  rate 
of  diffusion  the  longer  the  tube  the  greater  the  reduction  of  the 


From  a  photograph  by  the  author. 

FIG.   179. — Small  discharge  from  relief  valve  near  draft  tube  in  Cornell 
University  power  plant. 

kinetic  energy  of  the  water.  Therefore,  in  some  cases  it  is  de- 
sirable to  have  a  long  draft  tube  even  though  the  runner  might 
be  set  very  near  the  water  level. 

On  account  of  the  function  which  is  fulfilled  by  the  draft  tube  it 
is  properly  regarded  as  an  integral  part  of  the  turbine.  Con- 
sidering the  turbine  and  the  draft  tube  as  a  unit,  it  may  be  seen 
that  the  less  the  kinetic  energy  lost  from  the  mouth  of  the  tube 
the  higher  the  efficiency  of  the  wheel,  thus  justifying  the  state- 


188 


HYDRAULICS 


ment  of  the  preceding  paragraph.  But  it  is  not  yet  clear  just 
how  this  saving  in  the  draft  tube  enables  the  turbine  itself  to 
deliver  more  power  until  we  consider  the  effect  of  the  draft  tube 
upon  the  pressure  at  the  exit  from  the  runner.  The  less  the  losses 
within  the  draft  tube  and  the  less  the  discharge  loss  from  the 
draft  tube  the  less  the  pressure  may  be  at  this  point. 

130.  Cases  and  Settings. — The  turbine,  draft  tube,  and  all 
parts  intimately  connected  with  it  comprise  what  is  called  the 


From  a  photograph  by  the  author. 

FIG.   180. — Discharge  from  relief  valve  when  opened. 

setting.  Impulse  wheels  are  almost  always  set  with  horizontal 
shafts,  but  reaction  turbines  may  have  either-  horizontal  or 
vertical  shafts.  For  large  units  under  low  heads  the  vertical 
shaft  is  the  most  recent  practice,  as  it  permits  several  desirable 
features  to  be  attained.1  Occasionally  several  runners  may  be 

1  H.  B.  Taylor,  "  Present  Practice  in  Design  and  Construction  of  Hydraulic 
Turbines,"  Canadian  Soc.  of  C.  E.,  Jan.  15,  1914. 


DESCRIPTION  OF  THE  REACTION  TURBINE    189 

mounted  on  the  same  shaft  but  the  tendency  is  to  eliminate  such 
construction  and  have  larger  runners  and  fewer  of  them,  and  two 
on  the  same  shaft  as  in  Fig.  166,  is  as  many  as  are  desirable.  As 
in  the  case  of  the  impulse  wheel,  we  may  have  two  separate 
turbines  connected  to  a  single  generator. 


Courtesy  of  S.  Morgan  Smith  Co. 

FIG.  181. — Reaction  turbine  in  open  flume. 

Under  low  heads  of  not  more  than  20  or  30  ft.,  we  may  have 
the  open  flume  setting  such  as  is  shown  in  Fig.  181,  but  for  higher 
heads  this  is  not  practicable.  For  either  low  or  moderate  heads 
the  wheel  may  be  enclosed  within  a  concrete  case  as  in  Fig.  182, 


190 


HYDRAULICS 


but  the  action  of  the  turbine  is  no  different  from  that  in  the  pre- 
ceding case.  The  water  has  no  free  surface  immediately  above 
the  turbine  but  it  is  under  practically  the  same  pressure  as  if  it 


I  LOW  TAIL 
WATER 


IsLOTFOR 
T5TOP  LOGS 


Courtesy  of  S.  Morgan  Smith  Co. 

FIG.  182. — Reaction  turbine  in  concrete  case. 

did  have.  The  only  difference  is  that,  since  the  area  of  the 
water  passage  is  less  than  before,  the  velocity  with  which  the 
water  approaches  the  turbine  will  be  somewhat  higher,  and  thus 


DESCRIPTION  OF  THE  REACTION  TURBINE    191 


there  will  be  a  lesser  rate  of  acceleration  as  the  water  enters  the 
guide  vanes.  For  still  higher  heads  a  concrete  case  would  be 
unsuitable  and  then  the  guide  vanes  are  surrounded  by  a  metal 


I  LOW  TAIL 
WATER 


SLOT  FOR 
STOP  LOGS 


Courtesy  of  S.  Morgan  Smith  Co. 

FIG.   183. — Reaction  turbine  in  metal  case. 


case  as  shown  in  Fig.  183.  The  only  difference  between  this 
and  the  other  two  cases  so  far  as  the  hydraulics  is  concerned  is 
largely  one  of  appearances,  save  that  the  velocity  of  the  water  as 


192 


HYDRAULICS 


it  approaches  the  guides  may  be  somewhat  higher  owing  to  the 
smaller  area. 

In  order  that  the  water  may  have  the  same  velocity  of  ap- 
proach to  the  guides  all  around  the  circumference,  the  spiral  case 
is  frequently  used.  Cases  of  this  type  are  illustrated  in  Figs. 
175,  183,  184,  and  185.  In  Fig.  184  may  also  be  seen  the  main 
gate  valve  which  may  be  used  to  shut  off  the  water  more  com- 
pletely than  is  possible  with  the  wicket  gates,  and  on  the  right- 
hand  side  may  be  seen  a  portion  of  the  draft  elbow.  Very  large 


Courtesy  of  S.  Morgan  S^nith  Co. 

FIG.  184.  —  Spiral  case  turbine  showing  main  gate  valve,  shifting  ring  and 
guide  vanes,  and  draft  elbow. 


cases  are  built  in  sections  as  shown  in  Fig.  185.  The  spiral  case 
is  considered  the  most  desirable  type  though  other  less  expensive 
ones  are  sometimes  used. 

In  Fig.  186  is  shown  a  type  of  turbine  that  might  be  set  as  in 
Fig.  181,  while  in  Fig.  187  we  get  a  glimpse  into  the  intake  of  a 
large  turbine  set  as  in  Fig.  182.  In  such  a  setting  the  runner  and 
guide  vanes  may  be  surrounded  by  a  "speed  ring"  such  as  shown 
in  Fig.  188.  The  columns  which  support  the  upper  crown  plate 
and  its  load  are  made  of  a  shape  similar  to  guide  vanes  so  as  to 


DESCRIPTION  OF  THE  REACTION  TURBINE    193 


From  a  photograph  by  the  author 

FIG.  185. — Large  spiral  case  for  Canadian  Light  and  Power  Co.  in  shop  of 

I.  P.  Morris  Co. 


13 


194 


HYDRAULICS 


reduce  eddy  losses  and  also  to  give  the  water  the  proper  direction 
as  it  enters  the  real  guide  vanes.  In  Fig.  189  we  see  a  vertical 
shaft  turbine  for  a  higher  head. 

131.  Conditions  of  Service. — The  reaction  turbine  is  well 
adapted  for  service  under  low  heads  especially  for  large  powers. 
They  may  also  be  used  very  satisfactorily  for  heads  of  several 
hundred  feet.  The  highest  head  that  has  ever  been  employed 


Courtesy  of  Allis-Chalmers  Mfg.  Co. 


FIG.   186. — Vertical  open-flume   turbine  for  Eastern  Michigan  Edison  Co. 
h  =  14';  N  =  100;  Up.  =  575. 

for  a  reaction  turbine  is  670  ft.  for  two  6,000-hp.  units  installed 
by  the  I.  P.  Morris  Co.  in  Mexico.  There  are  several  cases  where 
heads  of  over  500  ft.  have  been  used. 

The  most  powerful  turbine  unit  yet  built  is  shown  in  Fig.  190, 
the  power  of  the  two  wheels  combined  being  25,000  hp.  The 
greatest  power  developed  in  a  single  runner  is  22,500  hp.  in  a 


DESCRIPTION  OF  THE  REACTION  TURBINE    195 


Courtesy  of  Mississippi  River  Power  Co. 

FIG.  187. — Intake  for  turbine  at  Keokuk. 


From  a  photograph  by  the  author. 

FIG.  188. — Speed  ring  for  Canadian  Light  &  Power  Co.  in  shop  of  I.  P. 

Morris  Co. 


196 


HYDRAULICS 


wheel  built  by  Allis-Chalmers  for  service  under  a  head  of  480  ft. 
It  is  a  double-discharge  runner,  however.  The  greatest  power 
ever  developed  in  a  single-discharge  runner  is  20,000  hp.  in  the 
runners  shown  in  Figs.  167  and  172. 

But  the  power  of  a  turbine  depends  not  only  upon  its  sbe  but 
also  upon  the  head  under  which  it  operates.     Thus  the  most 


Courtesy  of  S.  Morgan  Smith  Co. 

FIG.  189. — Vertical  shaft  spiral  case  turbine  for  Great  Falls,  Moiit. 
h  =  150';  N  =  200;  Hp  =  15,000. 

powerful  turbines  may  not  be  as  big  in  size  as  others  which 
develop  less  power  because  they  run  under  lower  heads.  The 
largest  runners  in  the  world  in  size  are  those  at  Cedars  Rapids, 
Canada,  one  of  which  is  shown  in  Fig.  170.  They  slightly  exceed 
in  size  those  at  Keokuk. 


DESCRIPTION  OF  THE  REACTION  TURBINE    197 


Courtesy  of  I.  P.  Morris  Co. 

FIG.  190.— Washington  Water  Power  Co.     Two  22,500  hp.  units  at  200 
r.p.m.  under  head  of  168  ft. 


CHAPTER  XII 
WATER  POWER  PLANTS 

132.  Elements  of  a  Water  Power  Plant. — A  complete  water 
power  development  may  comprise  a  great  deal  of  construction 
and  equipment  aside  from  the  power  house  and  contents,  so 


From  a  photograph  by  the  author. 

FIG.  191. — Penstock  leading  to  Drum  power  house  of  Pacific  Gas  &  Elec. 
Co.  under  1375  ft.  head. 

much  so  that  the  cost  of  the  latter  is  often  a  small  proportion 
of  the  total  investment.  For  a  complete  plant  some  or  all  of 
the  following  details  may  be  required  according  to  the  physical 
situation. 

198 


WATER  POWER  PLANTS 


199 


A  dam  of  some  sort  is  usually  essential.  It  may  be  nothing 
more  than  a  wing  wall  extending  a  short  way  into  the  river  to 
divert  a  small  portion  of  the  flow,  or  it  may  extend  clear  across 
the  stream.  In  the  latter  case  the  water  level  will  be  raised 
above  its  former  height  and  also  a  certain  amount  of  water  will 
be  stored  up  by  it.  If  the  contour  of  the  land  permits,  a  dam 
may  create  an  artificial  lake  or  storage  reservoir.  In  some  cases 
the  power  plant  draws  water  directly  from  this  body  and  in  other 
cases  it  would  be  used  merely  as  a  "  feeder." 

The  water  is  conducted  to  the  power  house  through  canals, 
flumes,  pressure  tunnels,  or  pipe  lines,  as  the  case  may  be.  It  is 


From  a  photograph  by  the  author. 

FIG.  192. — Pelton-Doble  impulse  wheels  in  Drum  power  house  of  Pacific 
Gas  &  Elec.  Co.  h  =  1375'  static  or  1300'  under  normal  load;  N  =360; 
Hp  =  8,500  per  wheel. 

not  uncommon  for  the  water  to  be  carried  from  5  to  10  miles  or 
more  in  order  to  permit  the  utilization  of  a  higher  fall  than  could 
be  obtained  near  the  intake.  It  is  desirable  that  the  water  be 
kept  at  as  high  an  elevation  as  possible  during  the  first  portion 
of  its  course  as  this  permits  the  use  of  open  channels  or  low-pres- 
sure pipes,  which  is  cheaper  than  if  the  water  had  to  be  carried 
under  high  pressure  all  the  way.  This  portion  of  the  conduit  is 
often  called  the  "flow-line"  from  the  fact  that  its  main  function 
is  to  deliver  water  and  not  to  transmit  pressure. 


200 


HYDRAULICS 


From  a  photograph  by  F.  H.  Fowler. 

FIG.  193. — Las  Plumas  plant  at  Big  Bend  on  the  Feather  River  in  Cali- 
fornia, containing  six  reaction  turbines  of  18,000  hp.  each  under  a  head  of 


WATER  POWER  PLANTS  201 

At  the  end  of  such  a  flow-line  the  water  will  be  abruptly 
dropped  down  the  hillside  as  shown  in  Fig.  191.  This  portion  of 
the  pipe  line  is  the  penstock. 

Where  the  distance  from  the  intake  to  the  power  plant  is  a 
number  of  miles,  it  is  desirable  that  there  be  some  break  in  the 
continuity  of  flow,  on  account  of  speed  regulation.  If  conditions 


From  a  photograph  by  the  author. 

FIG.  194. — San  Francisquito  Power  Plant  No.  1  on  the  Los  Angeles 
Aqueduct.  Static  head  from  maximum  water  level  in  surge  chamber  on 
creast  of  hill  to  the  nozzles  is  941  ft. 

permit,  a  forebay  may  be  constructed  at  the  head  of  the  penstock. 
The  forebay  is  a  reservoir  of  limited  capacity  whose  function 
is  to  equalize  the  flow.  Into  it  the  water  may  be  delivered  at  a 
uniform  rate,  while  from  it  the  water  may  be  drawn  by  the  pen- 
stock at  varying  rates  according  to  the  demands  upon  the  tur- 


202 


HYDRAULICS 


bines.     Thus  the  fluctuations  in  the  flow  of  water  through  the 
turbines  need  not  extend  back  all  the  way  to  the  source. 

Where  a  forebay  is  impossible  or  not  really  necessary,  it  is 
desirable  to  provide  surge  chambers  or  other  means  of  relieving 
the  abnormal  conditions  attendant  upon  changes  of  flow.  In 
the  upper  left-hand  corner  of  Fig.  193  is  seen  a  small  surge  cham- 
ber, and  an  overflow.  The  five  penstocks  receive  water  from  a 
pressure  tunnel  3  miles  in  length.  In  case  of  a  sudden  decrease 
in  discharge  through  the  turbines,  the  excess  water  could  surge 
up  the  large  pipe  line  running  up  the  hillside  and  if  the  surge  was 


From  a  photograph  by  the  author. 

FIG.  195.— Surge  chamber  designed  by  W.  F.  Durand.  It  is  100  ft.  in 
diameter  at  the  top  and  the  maximum  water  level  is  150  ft.  above  the  pres- 
sure tunnel.  Only  35  ft.  projects  above  the  ground. 


great  enough  some  water  would  overflow,  thus  preventing  any 
excessive  increases  in  pressure. 

In  Fig.  194  is  shown  a  power  plant  with  a  large  surge  chamber 
at  the  end  of  a  pressure  tunnel  which  is  7.76  miles  in  length  and 
in  Fig.  195  is  seen  as  much  of  the  surge  chamber  as  is  visible 
above  the  ground.  This  is  also  provided  with  a  spillway  so  that 
it  may  overflow  if  the  surge  is  violent. 

The  power  plant  shown  in  Fig.  196  receives  water  through  a 
conduit  1,711  ft.  in  length  and  is  equipped  with  a  large  air 
chamber  within  the  power  house  to  absorb  shocks. 

The  water  from  a  power  plant  may  be  discharged  directly  into 


WATER  POWER  PLANTS  203 

some  natural  stream  or  it  may  be  necessary  to  construct  an 
artificial  channel  for  a  tail  race,  as  in  Fig.  200.  In  other  cases, 
as  with  some  of  the  plants  at  Niagara  Falls,  the  tail  race  may  be 
a  long  tunnel. 

133.  High-head  Plants. — It  is  impossible  to  establish  any 
definite  number  of  feet  which  is  required  to  differentiate  a  high- 


From  a  photograph  by  the  author. 

FIG.  196. — Cornell     University    hydro-electric    plant.     Head  =  140    ft., 
1-550  hp.  turbine,  2-280  hp.  impulse  wheels,  2-50  hp.  impulse  wheels. 

head  from  a  medium-  or  a  low-head  plant.  A  high-head  is  one 
of  several  hundred  feet  or  more,  while  a  low-head  plant  would 
doubtless  be  under  50  ft.  But  one  type  shades  very  gradually 
into  the  other. 


204 


HYDRAULICS 


In  Fig.  197  is  shown  a  high-head  development,  where  a  fall  of 
4,000  ft.  is  divided  between  two  power  houses  in  series.  In  this 
one  view  may  be  seen  a  complete  plant  with  many  of  the  features 
that  have  been  described,  except  that  a  forebay  is  not  required. 
The  mountain  ranges,  which  rise  to  a  height  of  11,000  ft.,  provide 
a  water  shed,  the  runoff  from  which  is  gathered  by  a  lake  about 
4  miles  long,  and  with  an  elevation  of  6,000  ft.  The  lake  is 
created  by  the  erection  of  three  dams  which  can  be  seen  placed 
in  gaps  in  the  hills.  From  the  lake  the  water  flows  down  the 
penstocks  to  the  first  plant.  The  discharge  from  this  supple- 


Courtesy  of  Stone  and  Webster. 

FIG.  197. — Big  Creek  development  of  Pacific  Light  and  Power  Corpora- 
tion. The  fall  from  the  lake  to  the  first  power  house  is  2100  ft.,  and  from 
that  to  the  second  power  house  is  1900  ft. 

mented  by  some  water  from  a  little  stream  then  flows  through 
a  tunnel  for  a  way  until  it  takes  another  drop  to  the  second  power 
house  which  can  be  seen  in  the  lower  part  of  the  picture,  a  little 
to  the  left  of  the  center.  In  Figs.  198  and  199  are  closer  views  of 
these  two  plants.  At  the  upper  right-hand  corner  of  Fig.  198  can 
be  seen  two  standpipes,  the  water  level  in  which  will  be  nearly 
as  high  as  that  in  the  lake  so  that  the  entire  2,100  ft.  drop  is  shown 
here.  The  standpipes  for  the  second  plant  can  barely  be  seen 
on  the  crest  of  the  hill  in  the  upper  left-hand  corner  of  Fig.  199. 
A  high-head  plant  requires  but  little  water  for  a  given  amount 
of  power  and  it  is  usually  so  situated  that  a  storage  reservoir  is 


WATER  POWER  PLANTS 


205 


206 


HYDRAULICS 


Courtesy  of  Stone  and  Webster. 

FIG.  199.— Power  Plant  No.   2  at  Big  Creek. 

Up.  =  40,000. 


Static  head  =  1900  ft.; 


Courtesy  of  I.  P.  Morris  Co. 

FIG.  200. — Appalachin  Power   Co.   development   No.   2.     Head 
four  turbines  of  6000  hp.  each  at  116  r.p.m. 


49  ft., 


WATER  POWER  PLANTS 


207 


208  HYDRAULICS 

to  be  had.  Consequently  it  may  be  able  to  run  for  a  long  time 
merely  on  the  water  that  is  conserved  by  the  creation  of  such 
reservoirs.  It  is  always  necessary  to  have  a  penstock  and  many 
of  the  other  details  that  have  been  enumerated! 

134.  Low-head  Plants. — A  typical  low-head  plant  is  shown  in 
Fig.  200.  The  head,  under  which  the  turbines  operate,  has  been 
practically  created  by  the  erection  of  a  dam.  There  are  no  pipe 
lines  and  the  body  of  water  produced  by  the  dam  now  becomes 
the  forebay.  The  turbines  in  such  a  plant  may  have  any  one 
of  the  three  types  of  settings  shown  in  Figs.  181,  182,  and  183. 

It  may  be  seen  that  fluctuations  in  the  flow  of  the  river ;  with 
consequent  changes  in  water  level,  cause  variations  in  the  head 
under  which  the  turbines  operate.  This  is  something  that 
scarcely  exists  in  a  high-head  plant.  Also  low  heads  are  usually 
found  in  fairly  flat  countries,  where  the  nature  of  the  topography 
renders  it  impractical  to  store  up  large  quantities  of  water  and 
furthermore  under  a  low  head  a  large  amount  of  water  is  required 
to  develop  a  given  power.  This  makes  it  impossible  to  run  very 
long  on  storage  and  hence  the  plant  is  dependent  upon  a  regular 
stream  flow. 

The  differences  between  the  high-  and  low-head  plants  are 
such  as  to  require  turbines  of  different  characteristics  in  order  to 
meet  the  conditions  most  satisfactorily.1 

Another  typical  low-head  plant  is  shown  in  Fig.  201.  The 
length  of  the  dam  across  the  river  is  nearly  a  mile.  While  a 
low-head  plant  is  often  free  from  many  of  the  items  that  are 
required  in  a  high-head  development,  it  must  be  remembered 
that  it  must  be  built  to  handle  large  volumes  of  water  and  much 
massive  construction  is  required. 

1  R.  L.  Daugherty,  "Hydraulic  Turbines,"  Chap.  XII. 


CHAPTER  XIII 
THEORY  OF  THE  IMPULSE  WHEEL 

135.  Action  of  the  Water. — The  impulse  wheel  is  more  ac- 
curately described  as  a  tangential  waterwheel  from  the  fact  that 
the  center  line  of  the  jet  is  tangent  to  the  path  described  by  the 
center  of  the  buckets.  The  latter  is  called  the  " impulse  circle" 
and  computations  are  based  upon  the  linear  velocity  of  the  wheel 
at  this  radius.  For  an  impulse  wheel  the  nominal  value  of  D 
is  the  diameter  of  this  circle. 


FIG.  202. 

It  is  often  stated  that  the  jet  at  impact  is  also  tangent  to  this 
circle,  but  this  is  not  a  true  representation  of  the  facts  as  Fig. 
202  will  show.  The  jet  strikes  the  buckets  before  they  arrive 
at  a  point  directly  under  the  center  of  rotation  and  hence  the 
angle  A\  is  not  zero.  Observation  of  various  Pelton  wheels  in 
action  has  convinced  the  author  that  average  values  of  AI 
may  vary  from  5°  to  20°,  according  to  the  design  of  the  wheel. 
The  value  of  A\  must  be  an  average  value  for  a  given  wheel 
from  the  fact  that  the  bucket  moves  through  a  certain  angle 
from  the  time  it  first  enters  the  jet  until  the  last  drop  of  water 
has  struck  it. 

14  209 


210  HYDRAULICS 

The  illustration  also  shows  that  when  a  bucket  first  enters 
the  jet  it  cuts  off  the  water  from  the  preceding  bucket  and 
leaves  a  "slug"  of  water  to  catch  up  with  the  latter  and  to 
complete  its  work  upon  it.  Thus  the  water  may  be  acting  upon 
several  buckets  at  the  same  time.  This  explains  why  there  is  a 
difference  between  the  W  of  Art.  109  and  W.  The  former  is 
the  amount  of  water  acting  upon  a  single  bucket,  the  latter  is 
the  total  acting  upon  all  the  buckets.  It  is  not  necessary  to 
know  how  many  buckets  are  in  action  at  a  time  for,  since  the 
wheel  does  not  move  away  from  the  nozzle,  it  follows  that  all 
of  the  water  discharged  by  the  nozzle  may  act  upon  it. 

But  it  is  not  necessarily  true  that  the  wheel  utilizes  all  of 
the  water  under  all  circumstances.  Suppose  for  instance  that 
the  buckets  were  to  move  as  fast  as  the  jet;  it  would  then  be 
seen  that  none  of  the  water  could  overtake  them,  but  that  all  would 
go  right  on  through.  And  for  speeds  somewhat  less  than  this  a 
portion  of  the  water  would  deliver  its  energy  to  the  buckets 
and  the  latter  portions  of  the  intercepted  "slugs"  would  not 
be  able  to  overtake  the  buckets  before  they  had  swung  up  above 
the  line  of  action  of  the  jet.  The  problem  is  to  so  design  the 
buckets  and  the  wheel  that  all  of  the  water  in  the  jet  will  be  able 
to  do  its  work  upon  the  wheel,  when  running  at  the  proper  speed. 
For  speeds  much  above  the  normal  speed  a  certain  amount  of 
water  must  necessarily  go  right  on  through  without  having 
had  a  chance  to  do  work. 

By  the  proper  speed  or  normal  speed  is  meant  the  one  that 
the  wheel  should  have  for  the  jet  velocity  in  question.  A  high 
jet  velocity  would  require  a  high  wheel  speed  and  vice  versa. 
In  fact  we  are  concerned  with  the  relation  between  the  various 
velocities  rather  than  with  their  actual  values,  and  hence  it  is 
desirable  to  introduce  factors  which  shall  express  this  relation- 
ship and  be  independent  of  the  head.  Thus  if  the  jet  velocity 
be  denoted  by  Vi  and  the  linear  velocity  of  the  bucket  at  the 
impulse  circle  by  u\,  we  may  use  cv  and  0  such  that 

Vi  =  cvV2^h  (120) 

ui  =  <j)V2gh  (121) 

It  may  be  seen  that  cv  is  the  velocity  coefficient  of  the  nozzle, 
the  value  of  which  is  constant  for  any  setting  of  the  needle. 
Thus  for  any  given  value  of  <j>  the  relation  between  Vi  and  u\ 
is  known  at  once  regardless  of  the  value  of  h. 


THEORY  OF  THE  IMPULSE  WHEEL  211 


/  «*, 


FIG.  203. 


U\ 


Ml  V! 


«2=0 


V". 


0=0.0  0=0.20  0=0.45  0=0.55  0=0.80 

FIG.  204. — Velocity  diagrams  for  different  speeds. 


From  a  photograph  by  the  author. 

FIG.  205. — A  42-in.  Pelton-Doble  impulse  wheel. 


212 


HYDRAULICS 


From  a  photograph  by  the  author. 

FIG.  206. — Showing  discharge  from  buckets  when  wheel  is  at  rest,  or  <p  =  0.0. 


From  a  photograph  by  the  author. 
FIG.  207.— Wheel  running  at  slow  speed.     ?  =  0.20. 


THEORY  OF  THE  IMPULSE  WHEEL  213 


From  a  photograph  by  the  author. 
FIG.  208. — Wheel  running  at  normal  speed.     <p  =  0.45. 


From  a  photograph  by  the  author. 

FIG.  209. — Wheel  running  at  higher  speed.     <f>  =  0.55. 


214  HYDRAULICS 

The  absolute  path  of  the  water  and  the  velocity  vectors  at 
discharge  from  the  buckets  may  be  seen  in  Fig.  203.  For  dif- 
ferent wheel  speeds  under  the  same  head,  which  means  different 
values  of  </>,  we  should  have  such  diagrams  as  are  shown  in  Fig. 
204.  As  the  speed  of  the  wheel  increases  from  zero,  the  angle 
of  deflection  of  the  jet  continually  decreases.  It  may  also  be 
seen  from  the  diagrams  that  the  value  of  Vz  is  relatively  high 
when  the  wheel  is  at  rest,  that  it  becomes  a  minimum  at  such  a 


From  a  photograph  by  the  author. 
FIG.  210. — Wheel  at  run-away  speed.     <p  =  0.80. 

speed  that  A2  is  approximately  equal  to  90°,  and  then  increases 
again. 

The  action  of  the  water  as  just  described  is  illustrated  by 
some  rather  unusual  photographs  taken  of  a  42-in.  wheel  in 
action.  The  side  of  the  casing  was  removed  for  the  purpose. 
The  needle  was  withdrawn  as  far  as  possible  so  that  the  maximum 
size  jet  which  the  design  permitted  is  shown  in  the  photo- 
graphs. In  Fig.  206  the  wheel  was  prevented  from  rotating 
by  applying  a  sufficient  torque  to  the  shaft.  The  jet  cannot  be 
seen;  but  the  water  leaving  the  bucket  is  shown.  In  Fig.  208 
the  wheel  is  running  at  its  most  efficient  speed.  The  water 


THEORY  OF  THE  IMPULSE  WHEEL  215 

leaving  the  buckets  drops  down  into  the  tailrace  with  most  of 
its  energy  abstracted.  In  Fig.  210  the  wheel  is  shown  at  run- 
away speed,  all  load  having  been  removed  save  its  own  friction 
and  windage  and  that  of  the  generator  to  which  it  is  direct- 
connected. 

136.  Force  Exerted  by  Jet.  —  The  tangential  waterwheel, 
Pel  ton  wheel,  or  impulse  wheel,  as  it  is  variously  called,  is  really 
an  impulse  turbine  with  approximately  "  axial"  flow.  By  this 
is  meant  that  n  =  r2.  The  latter  is  not  strictly  true  but  is 
sufficiently  close  for  all  practical  purposes.1 

The  force  desired  is  really  the  tangential  component  of  the 
resultant  force.  This  may  be  obtained  by  computing  the 
tangential  component  of  the  AF  in  equation  (83),  or,  since  r\ 
and  r2  are  equal,  it  may  be  obtained  from  equation  (100).  The 
results  are  identical.  The  desired  component  is 

W 

P  =  —  •  (Fi  cos  Ai  —  F2  cos  A  2). 

t7 

The  values  of  F2  and  A  2  depend  upon  the  jet  velocity  and  the 
speed  of  the  wheel,  and  are  therefore  variable  and  unknown. 
It  is  desired  to  replace  them  in  terms  of  Fi  and  u\  and  wheel 
dimensions  which  may  be  supposed  to  be  known.  It  is  thus 
necessary  to  find  some  relation  between  the  velocities  at  entrance 
and  those  at  discharge.  The  equation  of  flow  between  these 
two  points  will  not  be  derived  here,  but  for  the  impulse  turbine 
it  becomes 

=  ^i2  +  u2*  -  wi2  (122) 


where  k  is  a  coefficient  of  loss  in  flow  over  the  buckets,  such  that 
the  head  lost  is  kvz2/2g. 

By  trigonometry  we  have  that,  in  general, 

1^2  =  yj  +  V  _  2VjUi  cos  A!  (123) 

If  this  value  of  v\  were  inserted  in  equation  (122)  we  should 
have  vz  expressed  in  terms  of  known  quantities.  But  before 
proceeding  further  we  shall  make  an  assumption  which  will 
simplify  the  resulting  expression  without  involving  any  great 
error,  though  it  must  be  remembered  that  our  result  will  not  be 
precisely  correct.  While  it  is  not  true  that  Ai  =  0°,  it  is  never- 

1  For  further  details  see  "Theory  of  the  Tangential  Waterwheel,"  by 
R.  L.  Daugherty,  Cornell  Civil  Engineer,  vol.  22,  p.  164  (1914).  Also  see 
"Hydraulic  Turbines,"  Chap.  VII. 


216  HYDRAULICS 

theless  small  enough  so  that*  its  cosine  differs  from  unity  by  only 
a  few  per  cent.  Hence  we  shall  assume  cos  AI  =  1.00.  This 
reduces  the  value  of  v\  to 

vi  =  Vi  —  Ui 

for  this  special  case.  Since  we  are  also  assuming  that  u\  =  w2, 
we  shall  have  from  equation  (122)  that 


(i  + 


From  the  vector  diagram  it  is  also  seen  that 
Vz  cos  A  2  =  u2  +  vz  cos  a2. 
Replacing  w2  by  u\  and  inserting  the  approximate  value  of 


we  have 

7==    = 

VI  +  k 


V2  cos  A2  =  ui  H  --  7==    =  (Vi  —  HI). 


Inserting  this  value  in  the  expression  for  P  we  have 

—  •> 


The  equation  just  derived  shows  that  P  decreases  as  the 
speed  of  the  wheel  is  allowed  to  increase.  This  is  what  we 
should  expect,  since  the  value  of  AF  decreases  as  shown  by 
Figs.  206  to  210.  Equation  (124)  is  apparently  the  equation 
of  a  straight  line  between  P  and  u\.  However,  the  factor  k 
is  not  strictly  constant  and,  as  has  been  shown,  the  value  of 
W  will  decrease  for  speeds  much  above  the  normal.  Likewise 
the  correct  equation  involves  A  i  and  this  also  changes  somewhat 
with  the  speed.1 

The  torque  exerted  by  the  water  upon  the  wheel  may  be 
obtained  by  multiplying  P  by  ri,  the  radius  of  the  impulse 
circle.  The  torque  which  the  wheel  can  deliver  is  somewhat 
less  than  this  because  of  bearing  friction  and  windage. 

Fig.  211  shows  the  performance  of  a  certain  wheel  at  different 

1  The  exact  equation  for  all  impulse  turbines  is 

D       TF|"T7  xcosaz      ,  -       -  "I 

=  —  [Vi  cos  At  -  i*t*,  -     *Y+k  vVi*  +  zW  -  ZViUi  cos  Al  J 

where  x  =  r*/r\.  For  the  Pelton  wheel  or  axial  flow  turbine  x  =  1;  for 
the  outward  or  inward  flow  Girard  impulse  turbines  it  is  more  than  or  less 
than  unity  respectively. 


THEORY  OF  THE  IMPULSE  WHEEL 


217 


speeds  under  a  constant  head.  The  variation  in  the  force  at 
zero  speed  is  due  to  changes  in  the  angle  AI  with  different  posi- 
tions of  the  buckets.  This  is  shown  for  one  nozzle  setting  only, 
though  it  exists  in  all. 

137.  Power  of  Wheel. — Since  power  is  the  product  of  P  and 
u\  or  T  and  co,  it  may  be  seen  that  it  is  zero  when  the  wheel  is 
at  rest,  though  the  torque  is  then  a  maximum,  and  it  is  also 
zero  when  the  speed  is  a  maximum  for  the  torque  is  then  zero. 


100 


70 


60 


£  40 


30 


10 


Head= 


65.5  Ft. 


100 


400 


R.P.M. 

FIG.  211. — Relation  between  torque  and  speed.1 


GOO 


The  maximum  power  will  be  obtained  for  some  speed  between 
these  two  extremes,  as  shown  by  Fig.  212. 

Since  for  a  given  head  and  nozzle  opening  the  power  input 
is  constant  regardless  of  the  speed  of  the  wheel,  it  follows  that 
the  efficiency  is  directly  proportional  to  the  power  developed. 
But  it  should  be  noted  that  the  power  delivered  in  the  water 
increases  with  the  nozzle  opening  so  that  the  needle  setting 
that  gives  the  largest  power  is  not  necessarily  the  most  efficient. 

138.  Speed. — From  equation  (124)  we  should  conclude  that 
P  would  become  zero  when  u\  =  Vi  or  when  </>  =  cvj  the  value 
of  which  would  .be  about  0.98.  Also  if  we  should  multiply  this 


1  From  the  test  of  a  24-in.  wheel  by  F.  G.  Switzer  and  the  author. 


218 


HYDRAULICS, 


equation  by  u\  the  power  would  be  seen  to  be  a  maximum  when 
Ui  =  0.5Fi.  But  equation  (124)  is  only  an  approximate 
representation  of  the  actual  facts.  Because  of  the  large  amount 
of  water  that  is  not  utilized  at  high  values  of  <£,  and  also  because 


3.48  Turns 


0  200    RpM     300  500 

FIG.  212. — Relation  between  power  and  speed. 


the  bearing  friction  and  windage  prevent  the  torque  from  ever 
being  reduced  to  zero,  the  actual  maximum  speed  attained  by 
the  tangential  waterwheel  is  such  that  <f>  is  approximately 
equal  to  0.80. 

In  like  manner  the  maximum  power,  and  hence  the  maximum 


THEORY  OF  THE  IMPULSE  WHEEL  219 

efficiency  for  a  given  nozzle  opening,  is  also  attained  when 
the  wheel  speed  is  something  less  than  0.5 V\.  Thus  in  actual 
practice  we  have  for  the  best  efficiency 

<(>e  =  0.43  to  0.48. 

In  practical  applications  we  are  usually  interested  in  the 
performance  of  a  wheel  at  a  constant  speed  under  a  constant 
head.  Values  for  this  may  be  obtained  from  Figs.  211  and  212 
by  following  along  any  vertical  line.  Generally  the  vertical 
line  should  be  the  one  for  the  speed  at  which  the  maximum 
efficiency  is  found.  The  resulting  curves  for  the  impulse  wheel 
would  be  very  similar  to  those  for  the  reaction  turbine  shown 
in  Fig.  216. 


FIG.  213. 

139.  Head  on  Impulse  Wheel. — The  nozzle  is  considered  an 
integral  part  of  the  impulse  wheel  and  hence  the  head  under 
which  the  wheel  is  said  to  operate  must  include  it.  If  C  in  Fig. 
213  indicates  a  point  at  the  base  of  the  nozzle, 

h  =  Hc  =  Pc  +  ~  (125) 

It  is  this  value  of  h  that  should  be  used  in  determining  the 
efficiency  of  the  wheel. 

140.  PROBLEMS 

1.  A  nozzle  having  a  velocity  coefficient  of  0.98  discharges  a  jet  6  in.  in 
diameter  under  a  head  of  800  ft.     This  jet  acts  upon  a  wheel  with  the  follow- 
ing dimensions:  diameter  6  ft.,  A\  =  10°,  a^  =  165°,  and  it  is  assumed 
k  =  0.70.     Find  the  force  exerted  upon  the  buckets  when  <f>  =  0.45. 

2.  Solve  problem  (1),  assuming  that  Ai  =0°. 

3.  Find  the  power  developed  upon  the  buckets  of  the  wheel  in  problem 
(1),  assuming  A\  =0°.     What  is  the  hydraulic  efficiency  of  the  wheel? 

4.  If  the  mechanical  efficiency  of  the  wheel  is  0.97,  what  is  the  gross 
efficiency  in  problem  (3)? 


220  HYDRAULICS 

6.  Assuming  A\  =0°  in  problem  (1),  find  the  power  lost  in  hydraulic 
friction  within  the  buckets.  Find  the  value  of  V%  and  determine  the  power 
carried  away  in  the  water  discharged  from  the  turbine. 

6.  Is  the  hydraulic  efficiency  of  an  impulse  wheel  dependent  upon  the 
head  under  which  the  wheel  is  run?     What  equation  would  express  the 
value  of  the  hydraulic  efficiency  for  any  tangential  waterwheel? 

7.  What  would  be  the  proper  r.p.m.  of  the  wheel  in  problem  (1)? 

8.  A  good  proportion  between  jet  and  wheel  is  that  the  diameter  of  the 
wheel  in  feet  should  equal  the  diameter  of  the  jet  in  inches.     Using  this 
ratio,  what  size  wheel  would  be  required  to  deliver  5,000  hp.  under  a  head 
of  1,400  ft.,  assuming  an  efficiency  of  82  per  cent.?     What  would  be  the 
speed  of  the  wheel? 

9.  The  maximum  speed  attained  by  the  wheel  of  Fig.  211  was  475  r.p.m. 
under  a  head  of  65.5  ft.     What  was  the  value  of  0? 

10.  The  best  efficiency  for  the  wheel  whose  curves  are  shown  in  Fig. 
212  was  found  with  the  needle  open  6  turns.     The  speed  was  275  r.p.m. 
and  the  head  65.5  ft.     What  was  the  value  of  <£e? 


CHAPTER  XIV 


^Stationary 


THEORY  OF  THE  REACTION  TURBINE 

141.  Introductory  Illustration. — The  reaction  turbine  is  so 
called  because  an  important  factor  in  its  operation  is  the  reac- 
tion of  the  streams  of  water  discharged  from  the  runner.  It 
is  well  to  bear  in  mind,  however,  that  the  total  dynamic  effort 
is  due  to  the  entire  change  in  the  momentum  of  the  water  just 
as  in  the  impulse  turbine. 

As  an  illustration,  consider  the  vessel  ABC  of  Fig.  214  into 
which  water  enters  across  AB  with  a  velocity  Vi  and  is  dis- 
charged at  C  with  a  velocity 
Vz'  Now  the  reaction  of  the 
jet  alone  could  be  determined  = 

2* 

by  an  application  of  Art.  110. 

But  the  total  force  is  due  not 

only  to  this  reaction  but  also 

to  the  impulse  of  the  water 

entering    at   AB.     It   is   not  pIG  2i4. 

feasible  to  separate  the  effect 

of  impulse  from  that  of  reaction,  neither  is  it  necessary  to  do 

so.     The  horizontal  component  of  the  total  dynamic  force  is 

obtained  directly  by 

W 
P  =  7(FIcosA1-F2). 

Suppose  now  that  this  vessel  moves  to  the  left  with  a  uniform 
translation  u.  Assume  that  in  some  way  the  water  is  still 
supplied  to  it  with  a  velocity  Vi.  This  might  be  the  case  if 
the  vessel  passed  under  a  series  of  stationary  passages  each  of 
which  in  turn  was  permitted  to  discharge  water  into  it.  The 
value  of  the  absolute  velocity  of  discharge  is  now 

Vz  =  u  —  vz. 
Inserting  this  value  above  we  have 

p  =  ~-  (Vi  cos  Ai  +  vt-  u)  (126) 

221 


222 


HYDRAULICS 


This  equation  indicates  that  P  decreases  as  the  speed  increases, 
just  as  in  the  case  of  the  impulse  turbine.  Also  the  water  enter- 
ing the  vessel  at  AB  is  under  pressure  and  is  not  a  free  jet. 
Therefore  Vi  must  be  less  than  \/2gh.  Since  all  the  passages 
are  completely  filled  with  water  the  equation  of  continuity  can 
be  applied,  and  it  will  show  that  V\  and  v2  are  inversely  pro- 
portional to  the  areas  of  their  respective  streams.  But  the  value 
of  ^2  depends  upon  the  losses  of  head,  and  in  a  real  turbine  these 


Values  of  0 
.0      0.1       0.2      0.3      0.4     0.5      0.6      0.7     0.8     0.9      1.0      1.1       1.2     1.3 


0         100       200 


400        500        600        700 
Revolutions  per  Minute 


900       1000      1100 


FIG.  215. — Test  of  27  in.     I.  P.  Morris  turbine.     Head  and  gate  opening 
constant.     Speed  variable.1 


hydraulic  losses  vary  with  the  speed.     Since  Vi  is  proportional 
to  vz,  it  follows  that  Vi  varies  with  the  speed  of  the  wheel. 

Thus  some  fundamental  differences  between  impulse  and 
reaction  turbines  are  that  in  the  former  Vi  =  cv\/2gh,  where 
cv  is  a  velocity  coefficient  nearly  equal  to  unity.  This  velocity, 
and  hence  the  amount  of  water  discharged  by  the  nozzle,  is 

1  Figs.  215  and  216  are  from  the  test  of  a  reaction  turbine  in  the  Cornell 
University  power  plant.  See  "Investigation  of  the  Performance  of  a 
Reaction  Turbine"  by  R.  L.  Daugherty,  Trans.  A.  S.  C.  E.,  vol.  78,  p.  1270 
(1915). 


THEORY  OF  THE  REACTION  TURBINE         223 

entirely  independent  of  the  design  of  the  wheel  and  its  operation. 
But  for  the  reaction  turbine 


(127) 

where  c  is  not  a  velocity  coefficient  but  a  factor  whose  value 
varies  from  about  0.6  to  0.8  for  ordinary  designs.  The  value 
of  c  depends  upon  the  design  of  the  wheel  and  the  speed  at  which 
it  is  run  under  a  given  head.  This  means  that  c  is  also  a 
function  of  <£,  where  <j>  has  the  meaning  given  by  equation  (121). 

With  the  radial-flow  type  of  turbine  centrifugal  force  also 
causes  the  value  of  c  to  vary  with  the  speed  of  the  wheel,  the  head 
remaining  constant.  The  centrifugal  force  opposes  the  flow 
of  water  in  the  case  of  the  inward-flow  turbine  so  that,  as  the 
speed  increases  under  a  constant  head,  the  discharge  tends  to 
decrease  as  shown  in  Fig.  215.  But  there  are  other  influences 
at  work  also,  so  that  for  some  inward-flow  turbines  the  value  of 
q  actually  increases  somewhat  as  the  speed  increases  above  zero, 
but  after  a  certain  speed  is  exceeded  the  rate  of  discharge  falls 
off  again. 

142.  Torque  Exerted.  —  The  preceding  article  merely  illustrates 
a  few  fundamental  points  regarding  the  reaction  turbine.  Since 
with  the  real  machine  the  radii  of  the  water  at  inflow  and  outflow 
differ  materially,  it  is  not  feasible  to  compute  the  force  exerted 
by  the  water  and  we  must  get  the  torque  instead.  Before  pro- 
ceeding any  further  with  the  theory  it  should  be  noted  that, 
while  our  equations  are  rational,  they  must  assume  that  all 
particles  of  water  move  in  similar  paths  with  equal  velocities. 
Actually  we  have  to  deal  with  average  values.  But  we  do  not 
know  these  average  values  with  any  precision.  For  example, 
we  have  no  assurance  that  the  angles  A\  and  a^  are  the  same  as 
the  angles  of  the  guide  vanes  and  the  runner  vanes  respectively. 
In  fact  we  have  some  evidence  to  indicate  that  they  differ  by 
as  much  as  5°  or  10°.  The  same  condition  exists  with  regard  to 
the  areas  of  the  streams  and  all  other  dimensions  used.  Thus 
the  numerical  results  of  such  computations  cannot  be  expected 
to  agree  precisely  with  actual  facts. 

Despite  this  the  theory  has  its  value.  It  serves  to  explain 
the  principles  of  operation  of  such  machines,  to  indicate  the  nature 
of  their  actual  characteristics,  and  to  account  for  numerous 
observed  facts.  In  design  the  theory  shows  what  proportions 
are  desirable  and  what  the  effect  of  certain  changes  of  dimensions 


224  HYDRAULICS 

would  be.  Thus  if  we  have  some  actual  test  data  to  work  from, 
the  theory  would  enable  us  to  alter  existing  designs  with  some 
degree  of  assurance. 

In  order  to  compute  the  torque  exerted  upon  the  runner  by 
the  water  we  should  take  the  fundamental   formula  of  Art.  113, 

W 

T  =  —  (r,Fi  cos  Al  -  r2F2  cos  A,). 
y 

Just  as  in  the  case  of  the  impulse  turbine  in  Art.  136,  the  values 
of  Vz  and  A2  are  variable  and  unknown,  and  it  is  necessary 
to  replace  them  in  terms  of  known  quantities.  It  is  assumed 
that  all  the  dimensions  of  the  wheel  and  the  values  of  Vi  and 
Ui  are  known. 

From  the  vector  diagram  it  may  be  seen  that 

Vz  COS  Az   =   Uz  +  Vz  COS  dz> 

But  Uz  =  (TZ/TI)UIJ  and,  since  the  passages  are  completely 
filled  with  water  in  the  reaction  turbine,  the  equation  of  con- 
tinuity gives  q  =  F{Vi  =  fzVz,  or 

vz  =  (Fi/fzWi  (128) 

(Contrast  this  procedure  with  that  for  the  impulse  turbine  in 
Art.  136  and  note  that  equation  (122)  does  not  apply  here.) 
Making  the  proper  substitutions  we  easily  derive 


T  =    ~  n[  (cos  A!  -  ^  •  jl  cos  a,)  V,  -  (~)  V]      (129) 


In  the  use  of  equation  (129)  we  should  have  to  determine  the 
value  of  W  for  any  value  of  u\,  either  by  experiment  or  by 
computing  the  rate  of  discharge  by  theory. 

143.  Power. — The  power  developed  by  the  water  is  deter- 
mined by  multiplying  T  by  the  angular  velocity.  The  torque 
actually  exerted  by  the  shaft  and  the  power  delivered  are  ob- 
tained by  multiplying  these  values  by  the  mechanical  efficiency. 

The  hydraulic  efficiency  of  the  turbine  is  obtained  by 

Tco       Wh"       h" 

eh  = 


Wh~    Wh         h 

It  is  difficult  to  obtain  the  hydraulic  efficiency  by  test  as  it 
is  necessary  to  determine  the  bearing  friction  and  also  the 
disk  friction  due  to  the  drag  of  the  runner  through  the  water 


THEORY  OF  THE  REACTION  TURBINE 


225 


in  the  clearance  spaces.     But  these  losses  may  be  allowed  for 
and  the  hydraulic  efficiency  then  secured  approximately. 

Because  of  the  necessary  defects  of  the  theory,  the  hydraulic 
efficiency  may  be  assumed  with  less  error  than  is  usually  in- 
volved in  computing  T.  For  turbines  of  rational  design  and 
running  at  their  proper  speeds  the  value  of  the  hydraulic  ef- 
ficiency may  range  from  0.80  to  0.95.  The  higher  values  are 
found  only  in  large  turbines  and  with  favorable  proportions. 
Only  experience  can  enable  one  to  select  the  proper  value  be- 


100 


70 


I- 


_800 


jpTOO 


500 


400 


no 


ytf 


30 


20 


10 


300 


200 


600  R. 


'.M. 


100 


100 


200        300        400 
B.HP.  =HP.  Output 


500 


FIG.  216. — Test  of  27  in.  I.  P.  Morris  turbine.     Head  and  <f>  approximately 
constant.     Gate  opening  variable. 


tween  these  two  extremes,  which  are  not  necessarily  limits. 
For  improper  speeds  and  incorrect  designs  no  values  can  be 
assigned. 

The  curves  of  Fig.  215  show  the  characteristics  of  a  reaction 
turbine  with  a  fixed  gate  opening  and  the  speed  variable. 
These  are  similar  to  those  of  the  impulse  wheel  except  that  q 
is  not  a  constant.  Hence  maximum  power  developed  and  the 
maximum  efficiency  do  not  necessarily  occur  at  the  same  speed. 
The  characteristics  of  the  same  turbine  at  a  constant  speed  are 

15 


226  HYDRAULICS 

shown  in  Fig.  216.     The  maximum  efficiency  is  88  per  cent,  at 
550  hp.  under  a  head  of  141.8  ft. 

144.  Speed. — Although  the  water  flowing  through  the  runner 
of  a  reaction  turbine  is  entirely  confined,  the  velocity  under- 
goes changes   similar  to  that  in  the  impulse  turbine,   except 
that  for  a  fixed  gate  opening  the  angle  Ai  is  constant.     Hence 
the  values  of  F2  and  A2  vary  in  just  the  same  way  as  is  shown 
by  Figs.  204  to  210. 

The  speed  at  which  the  efficiency  is  the  highest  will  be  some- 
where in  the  neighborhood  of  the  one  for  which  the  discharge 
loss  is  the  least.  The  value  of  F2  will  be  a  minimum  in  such  a 
case  as  Fig.  208.  For  this  condition  it  will  be  found  that  ap- 
proximately A  2  =  90°  or  that  uz  =  v^.  Note  that  these  two 
conditions  are  not  identical  but  they  differ  but  little  It  is 
customary  to  assume  one  or  the  other  according  to  convenience. 

In  the  case  of  the  reaction  turbine  Vi  is  less  than  for  an  im- 
pulse turbine  under  the  same  head.  But  the  water  at  entrance 
is  under  pressure  and,  as  it  flows  through  the  runner,  this  is 
converted  into  velocity.  Hence  at  discharge  v2  may  easily 
be  as  large  as  in  the  case  of  the  impulse  wheel.  And  if  u^  =  v2, 
it'may  be  seen  that  it2  will  be  about  the  same  in  either  type. 
But  with  the  inward  flow  reaction  turbine  u\  is  greater  than 
uz}  and  therefore  the  peripheral  velocity  of  the  reaction  turbine 
is  greater  than  that  of  the  impulse  wheel. 

Not  only  is  the  peripheral  speed  higher  for  maximum  efficiency 
but  also  the  runaway  speed  is  higher.  The  maximum  value  of 
<£  for  the  reaction  turbine  is  about  1.30,  though  with  some  it 
may  easily  exceed  this  value.  And  for  the  normal  speeds  at 
which  the  maximum  efficiency  is  obtained  we  have 

<f>e  =  0.60  to  0.90, 

the  exact  value  for  a  given  wheel  depending  upon  its  design. 

145.  Values   of   ce  and    4>e  for   Maximum   Efficiency. — The 
turbine  should  run  normally  at  such  a  speed  under  any  head  that 
the  maximum  efficiency  will  be  attained.     It  will  be  assumed 
that  this  speed  is  such  that  the  discharge  is  " radial"  or  A2  =  90°. 
The  angle  which  the  runner  vane  at  entrance  makes  with  u\ 
will  be  denoted  by  a'\.     As  the  turbine  is  ordinarily  designed, 
the  value  of  the  vane  angle  would  be  such  that  it  would  agree 
with  a\  as  determined  by  the  vector  diagram  for  this  same 
speed.     But  at  any  other  speed  the  value  of  ai  would  be  different 


THEORY  OF  THE  REACTION  TURBINE         227 

from  a'i,  hence  there  would  be  an  abrupt  change  in  the  direction 
of  the  water  entering  the  runner  giving  rise  to  what  is  known 
as  ''shock  loss." 

The  following  expressions  therefore  apply  only  to  the*  special 
case  where  A2  =  90°  and  a'\  =  a\.  From  the  vector  diagram 
of  velocities  we  have 

Vi  sin  A\  =  Vi  sin  a\  =  Vi  sin  a'\ 
Vi  cos  Ai  =  ui  +  Vi  cos  ai  =  HI  +  Vi  cos  a'i 
Eliminating  Vi  between  these  two  equations  we  have 

sin  (a'i  -  Ai)T7  f 

Ui=          sin  a',     ~7" 

as  the  relation  between  MI  and  Vi  when  there  is  no  loss  at  entrance 
to  the  runner. 

The  power  delivered  by  the  water  to  the  runner  may  be 
expressed  as 

Tco  =  Wh"  =  ehWh, 

where  T  has  the  value  given  by  equation  (100).     Thus 
Wh"  =  ~  (r,7i  cos  Ai  -  r2y2  cos  A2)^- 

If  the  discharge  is  "  radial,"  A2  =  90°  and  hence  Vz  cos  A2  =  0. 
Therefore 


Solving  equations  (130)  and  (131)  simultaneously,  we  have 


=     ,~n2gh  sin  a'i 


2      sin  (a'i  —  Ai)  cos 


eh  2gh  sin  (a'i  — 

Ui  = 


2       sin  a'i  cos  A! 
From  this  it  follows  that 


/  eh  sin  a'i 

:  \2sin  (a'i  -  Ai)cosAi 

fat  sin  (a'!  -~Ai) 
0e  ==  V2sina'iCOsA7 

It  must  be  borne  in  mind  that  equations  (132)  and  (133)  can  be 


228  HYDRAULICS 

applied  only  for  the  special  case  stated.  For  any  other  speed  a 
different  procedure  would  be  necessary  but  it  will  not  be  given 
here.1 

The  speed  desired  is  the  one  for  which  the  gross  efficiency 
is  a  maximum  and  this  may  not  be  quite  the  same  as  the  one  for 
which  the  hydraulic  efficiency  is  the  highest.  Hence  the  true 
value  of  (j>6  may  differ  slightly  from  the  value  given  by  equation 
(133). 

These  equations  appear  to  be  independent  of  conditions  at 
outflow  from  the  runner.  But  it  must  be  noted  that  they  are 
to  be  used  only  upon  the  assumption  that  the  dimensions  used 
at  exit  will  be  such  as  to  make  A2  =  90°,  when  ce  and  <j>e  have  the 
values  given. 

An  interesting  result  may  be  obtained  by  multiplying  equations 
(132)  and  (133).  This  gives 


This  would  indicate  that  all  other  things  being  equal,  the  higher 
the  value  of  </>e  the  smaller  the  value  of  ce.  With  the  impulse 
turbine,  to  which  these  equations  apply  also,  <f>e  is  small  but 
c  =  cv  and  is  near  unity.  With  the  reaction  turbine  <f>e  is  larger 
than  for  the  impulse  wheel  but  c  is  smaller. 

146.  Theory  of  th,e  Draft  Tube.—  If  the  draft  tube  is  properly 
designed  its  area  next  to  the  runner  should  be  such  that  the 
velocity  in  it  is  equal  to  72,  the  absolute  velocity  of  discharge, 
otherwise  there  will  be  an  abrupt  change  of  velocity  involving 
losses.  For  Fig.  217  we  may  write 

H2  =  p2  +  z2  +  F22/2fif,     and     H4  =  0. 

The  losses  between  points  (2)  and  (4)  are  made  up  of  the  friction 
losses  within  the  tube,  H'  '/,  and  the  discharge  loss  at  (3).  Apply- 
ing the  general  equation  between  points  (2)  and  (4)  we  have 

F92  Fs2 

^  =  ~  z*  ~  -%  +  H'f  +  %  (135) 

The  larger  the  diameter  of  the  tube  at  its  mouth  the  less  will 
be  the  value  of  Vs  and  hence  the  less  the  pressure  at  the  point  of 
discharge  from  the  runner.  But  if  too  great  a  rate  of  diffusion 
is  provided  for  in  the  tube  the  flow  in  it  will  be  unstable  and 
1  A  general  relation  between  c  and  <£  for  all  conditions  will  be  found  in 
the  author's  "Hydraulic  Turbines,"  Chap.  VIII. 


THEORY  OF  THE  REACTION  TURBINE 


229 


the  friction  loss  H' /  will  be  increased.  The  pressure  at  the  top 
of  the  draft  tube  should  not  be  made  less  than  about  5  ft.  absolute, 
and  the  value  of  z%  determined  accordingly.  A  " high-speed" 
turbine  runner  with  a  large  value  of  Vz  cannot  be  set  as  far  above 
the  water  level  as  a  " low-speed" 
turbine  with  a  smaller  value  of 

If  the  losses,  #'/,  be  assumed 
equal  in  the  two  cases  it  may  be 
seen  that  the  diverging  draft  tube 
increases  the  head  utilized  by  the 

turbine  by  |^1  —  \^rj  J^T* 

147.  Head  on  Reaction 
Turbine. — For  a  reaction  turbine 
the  draft  tube  is  an  integral  part 
of  the  machine,  hence  (Fig.  218) 
the  head  under  which  it  operates 
is  given  by 

h  =  Hc  -  Hf  =  pc  +  *c  +  ~-  (136) 

This  is  the  value  of  h  upon  which  computations  are  based,  and 
it  is  the  one  to  be  used  in  determining  the  efficiency  of  the 
turbine. 


FIG.  218. 

However,  though  the  turbine  maker  usually  constructs  or 
designs  the  draft  tube  also,  he  is  often  limited  by  the  conditions 
of  the  setting  and  may  not  be  able  to  use  the  proper  proportions. 
In  order  to  eliminate  this  defect  in  the  setting,  over  which  he 
has  no  control,  the  velocity  head  at  (E)  is  sometimes  deducted 
from  the  value  given  by  equation  (136).  If  it  were  feasible  to 


230  HYDRAULICS 

eliminate  the  friction  in  the  draft  tube  as  well  we  should  then 
have  the  efficiency  of  the  runner  alone,  which  is  independent,  of 
the  draft  tube.  But  what  we  usually  desire  is  the  efficiency  of 
the  entire  unit  from  the  intake  of  the  casing  to  the  tailrace. 

148.  PROBLEMS 

1.  A  certain  reaction  turbine  was  found  by  actual  test  to  have  a  hydraulic 
efficiency  of  0.83  when  0  =  0.670  and  c  =  0.665.     The  angles  were:  Ai  = 
13°  and  o'i  =  115°.     Compute  the  values  of  <j>e  and  ce  and  compare  with  the 
actual  values.     (The  slight  discrepancy  between   the  two  is  largely  due 
to  the  fact  that  shockless  entrance  and  radial  discharge  were  not  obtained 
at  exactly  the  same  speed.) 

Ans.  <f>e  =  0.678,  ce  =  0.628. 

2.  For  a  reaction   turbine  the  dimensions  are:  A\  =  35°,   a'i  =  136°, 
eh  =  0.845.     Compute  the  values  of  <t>e  and  ce. 

Ans.  <t>e  =  0.85,  ce  =  0.60. 

3.  In  the  test  of  the  Cornell  University  turbine  the  pressure  was  read 
by  a  mercury  manometer  attached  near  the  intake  flange  where  the  diameter 
was  30  in.     At  full  load  when  the  discharge  was  44.5  cu.  ft.  per  sec.,  the 
manometer  read  9.541  ft.  of  mercury,  the  top  of  the  shorter  mercury  column 
being  0.500  ft.  above  the  intake.     If  the  elevation  of  the  intake  above  the 
water  level  in  the  tailrace  is  9.230  ft.  find  the  head  on  the  turbine. 

Ans.  140.5ft. 

4.  In  the  turbine  of  problem  (3)  the  diameter  of  the  draft  tube  at  the 
upper  end  is  24.5  in.  and  at  the  bottom  it  is  42  in.     Find  the  gain  in  head 
due  to  its  use  when  the  discharge  is  44.5  cu.  ft.  per  sec. 

6.  The  top  of  the  draft  tube  in  problem  (4)  is  10.0  ft.  above  the  level  of 
the  water  in  the  tailrace.  Neglecting  the  friction  in  it,  but  considering  the 
discharge  loss  at  the  bottom,  find  the  pressure  at  its  top. 

6.  A  reaction  turbine  by  test  was  found  to  discharge  31.8  cu.  ft.  of  water 
per  sec.  when  running  at  600  r.p.m.  under  a  head  of  143.1  ft.  If  FI  =  0.535 
sq.  ft.  and  D  =  27  in.,  find  values  of  c  and  <£. 


CHAPTER  XV 
TURBINE  LAWS  AND  FACTORS 

149.  Operation  under  Different  Heads. — In  the  entire  dis- 
cussion in  the  two  preceding  chapters  we  have  assumed  that 
the  head  remained  constant  though  the  other  quantities  might 
vary.  But  a  turbine  may  be  installed  in  a  plant  where  the  head 
changes  from  time  to  time,  and  also  a  given  design  of  turbine 
might  be  used  in  different  plants  under  a  wide  range  of  heads. 
Thus  we  desire  to  investigate  this  p>hase. 

Let  us  recall  the  expression,  u\  =  <j>\/2gh.  Suppose  now  that 
a  turbine  is  compelled  to  run  at  a  constant  speed  while  the 
head  varies.  It  is  clear  that  0  also  varies  then  just  as  it  would 
in  the  preceding  case.  But  it  would  be  possible  under  some 
circumstances  to  chan  ge  the  speed  as  well  in  such  a  way  as  to 
keep  0  a  constant.  Hence  we  need  to  consider  two  distinct 
cases  when  the  head  changes ;  one  is  where  <j>  is  constant,  and  the 
other  is  where  <f>  also  changes. 

If  <j>  remains  constant,  the  wheel  speed  must  vary  as  \/h. 
But  a  definite  value  of  0  is  accompanied  by  a  definite  value 
of  c.  Hence  the  rate  of  discharge  must  also  vary  as  \/Ti,  since 
Vi  —  c\^2gh.  Now  the  power  of  the  water  is  proportional  to 
the  product  of  q  and  h.  Since  q  varies  as  \//^  it  follows  that 
the  power  varies  as  ffi.  And  in  similar  fashion  it  may  be  shown 
that  the  torque  varies  as  h. 

The  hydraulic  efficiency  is  a  function  of  c,  0,  and  the  turbine 
dimensions.  As  long  as  0  remains  constant  the  hydraulic 
efficiency  remains  the  same  regardless  of  the  head.  This  must  be 
true  because  the  hydraulic  losses  may  all  be  shown  to  vary  as  hP, 
just  as  the  power  input.  But  the  friction  of  the  bearings  and 
the  windage  or  the  disk  friction  of  the  runner  do  not  vary  in 
the  same  way.  It  is  not  possible  to  formulate  an  exact  law  for 
this  but  they  may  be  said  to  vary  between  N  and  N2.  Since 
N  varies  as  \/^  they  must  vary  between  h^  and  h.  Hence  the 
mechanical  losses  become  a  smaller  percentage  of  the  total  as 

231 


232  HYDRAULICS 

the  head  increases.1  But,  except  for  very  low  heads,  the  dif- 
ference in  the  efficiency  is  usually  a  matter  of  not  more  than  2 
or  3  per  cent,  at  most.  See  Fig.  219. 

Now  if  the  speed  remains  constant  while  the  head  changes,  or 
if  it  does  not  vary  as  \/Ji,  the  value  of  <f>  will  change.  Re- 
ferring to  Fig.  215,  it  may  be  seen  that  this  means  a  change 
of  c  also.  Hence  the  efficiency  will  change.  Thus  none  of  the 
simple  proportions  that  have  just  been  stated  will  be  true  in 
such  a  case.  It  is  impossible  to  calculate  the  new  results  unless 


Hydraulic  Efficiency 


Gross  Efficiency 


0  =  Constant 


Head 
FIG.  219. — Effect  of  head  upon  efficiency  of  a  given  turbine. 

curves,  such  as  those  of  Fig.  215,  are  available,  or  unless  we 
have  some  complex  equations  which  will  give  values  of  all  these 
quantities  for  any  value  of  <f>. 

150.  Different  Sizes  of  Runner. — If  a  series  of  runners  are 
all  built  of  the  same  design  with  the  same  angles  and  proportions 
so  that  one  is  simply  an  enlargement  or  reduction  of  another,  they 
should  all  have  the  same  values  of  <f>e  and  ce.  Since  their  peri- 
pheral speeds  would  all  be  the  same  under  a  given  head  it  follows 
that  their  rotative  speeds  would  be  inversely  proportional  to 
their  diameters.  And  the  area  F\  would  be  proportional  to  D2. 
Hence  their  capacity  and  power  would  vary  as  D2.  Thus  if  the 
performance  of  one  runner  is  known,  that  of  the  rest  of  the  series 

1  An  impulse  wheel  should  be  set  with  sufficient  space  on  either  side  of 
the  buckets  at  discharge  so  that  the  water  rebounding  from  the  walls 
will  not  strike  them.  The  velocity  with  which  the  water  rebounds  is  pro- 
portional to  V2  and  hence  to  \/h.  Therefore  if  this  space  is  not  ample 
for  all  values  of  h,  a  point  may  be  reached  where  this  action  would  decrease 
the  efficiency. 


TURBINE  LAWS  AND  FACTORS  233 

may  be  predicted  with  some  assurance,  due  allowance  being 
made  for  slight  increases  in  efficiency  as  the  size  increased. 
We  may  express  these  statements  algebraically  as  follows: 


N.  -  (137) 

where  1,840  is  a  constant  which  we  obtain  when  we  solve  for  N 
in  terms  of  peripheral  speed,  the  latter  being  given  by  equation 
(121).  For  the  two  types  of  turbines  in  common  use  we  have: 

Impulse  wheel  <f>e  =  0.43  to  0.48 

Reaction  turbine      <f>e  =  0.60  to  0.90 

according  to  design.     And  as  to  capacity 

q  =  K^Vh  (138) 

where  K\  has  the  following  range  of  values: 

Impulse  wheel          KI  =  0.0002  to  0.0005 
Reaction  turbine      KI  =  0.0014  to  0.0360 

It  must  be  understood  that  these  constants  are  based  upon  values 
corresponding  to  4>e  and  that  the  speed  of  the  wheel  must  be  such 
that  </>e  is  obtained  if  they  are  to  apply. 

Making  a  suitable  allowance  for  the  efficiency  the  power 
delivered  by  the  turbines  can  be  determined  when  the  discharge 
is  known. 

It  may  be  seen  that  the  peripheral  speed  of  the  reaction  turbine 
is  higher  than  that  of  an  impulse  wheel  and  that  it  may  be  varied 
through  a  wider  range  by  changes  in  the  design.  Also  the  values 
of  KI  show  that  for  a  given  diameter  a  reaction  turbine  can  dis- 
charge more  water,  and  hence  develop  more  power,  than  an  im- 
pulse turbine.  That  means  that  if  they  are  to  deliver  the  same 
power  the  diameter  of  a  reaction  turbine  will  be  much  less  than 
that  of  a  corresponding  impulse  wheel.  Thus  for  a  given  head 
and  power  the  rotative  speed  of  the  reaction  turbine  will  be  much 
higher  than  that  of  the  impulse  wheel  due,  both  to  its  higher 
peripheral  speed  and  to  its  much  smaller  size. 

151.  Specific  Speed.  —  A  useful  factor  in  turbine  work  is  one 
that  will  now  be  derived.  It  involves  the  head,  speed,  power,  and 
efficiency. 


234  HYDRAULICS 

Since  power  is  proportional  to  Z>2  and  to  h^  we  may  write, 
b.hp.  =  KzD2h3/*.     This  may  be  rewritten  as 


Inserting  the  value  of  D  as  given  by  equation  (137)  we  have 


1,8400,     ft        Vb.hp. 
Ne  h* 

Rearranging  this  and  letting  \/rK2  l,840<£e  =  N8,  we  have 


While  any  value  of  N  might  be  used,  the  expression  has  but 
little  meaning  unless  a  particular  value  is  employed.  That  is 
generally  understood  to  be  Ne,  the  value  of  N  at  which  the 
maximum  efficiency  under  a  given  head  is  attained.  As  to  the 
value  of  b.hp.  it  should  logically  be  the  one  for  which  the  maxi- 
mum efficiency  is  obtained  under  the  given  head.  But  in 
some  cases  the  value  of  the  maximum  power  at  this  same  speed 
is  used. 

The  quantity  Na  is  generally  known  as  specific  speed.  Other 
names  applied  to  it  are  unit  speed,  type  characteristic,  and 
characteristic  speed.  Its  value  indicates  the  class  to  which  a 
turbine  belongs.  Thus  we  have  seen  that  for  a  given  head  and 
power  the  impulse  wheel  runs  at  a  relatively  low  r.p.m.  There- 
fore it  would  have  a  low  value  of  N8. 

For  an  impulse  wheel  under  a  given  head  at  a  given  speed 
the  power  would  increase  with  the  size  of  the  nozzle  used. 
Thus  there  need  not  be  any  lower  limit  to  the  value  of  Na  but 
the  upper  limit  would  be  the  one  for  which  we  had  the  maximum 
size  jet  that  could  be  employed.  We  find  that  the  efficiency  is 
nol^  appreciably  reduced  until  after  we  pass  a  value  of  about 
4.5  for  Na  and  after  a  value  of  6  the  jet  is  so  large  for  the  size  of 
the  wheel  that  the  efficiency  drops  off  materially.  But  any 
value  above  4.5  involves  some  sacrifice  of  efficiency. 

For  the  reaction  turbine  we  have  limits  in  both  directions 
as  indicated  below,  though  these  may  be  extended  in  future 
designs.  The  values  of  the  specific  speed  are 


TURBINE  LAWS  AND  FACTORS  235 

Impulse  wheel  N8  =  0  to  4.5  (6  max.) 

Reaction  turbine       N8  =  10  to  100. 

For  a  given  turbine  the  value  of  Na  is  naturally  a  constant, 
but  it  is  also  practically  constant  for  a  whole  series  of  runners 
of  :the  same  design  regardless  of  size.  The  larger  the  diameter 
of  a  runner  the  greater  its  power  but  the  less  the  value  of  N  for  a 
given  head.  Hence  the  product  remains  constant. 

Values  of  N8  given  for  the  impulse  wheel  are  for  a  single  jet 
upon  a  single  wheel.  When  two  or  more  jets  are  used  the  power 
is  naturally  increased  without  changing  the  speed.  This  enables 
values  between  6  and  10  to  be  obtained,  if  necessary.  For 
values  above  100  the  conditions  are  impossible.  Either  the 
power  or  the  speed  of  the  unit  must  be  decreased. 

The  specific  speed  factor  shows  that  the  impulse  wheel  is  a 
low-speed,  low-capacity  turbine  and  the  reaction  turbine  is  a 
high-speed,  high-capacity  turbine.  The  use  of  these  words  is 
relative  rather  than  absolute.  Thus  the  turbine  in  Fig.  170 
runs  at  only  55.6  r.p.m.  but  its  specific  speed  is  82.3,  thus  in- 
dicating that  it  is  a  high-speed  wheel.  For  the  speed  is  high 
as  compared  with  that  of  other  turbines  of  the  same  power 
under  that  head.  For  instance  the  speed  of  an  impulse  wheel 
for  similar  conditions  would  be  only  4  r.p.m.  And  the  specific 
speed  of  the  highest  head  impulse  wheels  in  the  world  (Art. 
125)  is  only  0.592  though  they  run  at  500  r.p.m.  But  a  slow- 
speed  reaction  turbine  under  the  same  conditions  would  run  at 
8,450  r.p.m.  at  least,  and  a  high-speed  reaction  turbine  such  as 
those  at  Cedars  Rapids  would  run  at  69,300  r.p.m.  Of  course 
these  values  are  absurd  and  simply  demonstrate  the  fitness  of 
each  type  for  its  own  field. 

152.  Uses  of  Specific  Speed. — The  values  of  Ns,  as  of  all 
other  factors  in  this  chapter,  are  supposed  to  be  obtained  from 
test  data,  not  computed  by  theory.  They  serve  to  classify 
a  turbine  and  indicate  to  what  type  it  belongs.  They  are  useful 
in  selecting  units  for  a  prospective  plant.  For  such  a  case  the 
head  is  known  but  the  size  and  speed  of  the  units  is  not.  If  it 
is  desired  to  use  wheels  of  a  certain  type,  that  fixes  the  value  of 
N8  between  narrow  limits,  and  it  is  easy  to  compute  the  com- 
binations of  speed  and  power  that  can  be  produced.  Or,  if 
the  speed  and  power  be  fixed,  it  may  at  once  be  found  what 
type  of  turbine  would  be  required. 


236  HYDRAULICS 

153.  Factors  Affecting  Efficiency. — The  efficiency  of  the  im- 
pulse wheel  is  practically  independent  of  the  size  of  the  wheel. 
The  author  makes  this  statement  after  testing  sizes  from  12  in. 
to  84  in.  in  diameter  and  comparing  all  the  other  test  data  which 
is  accessible.  It  would  seem  reasonable  that  this  should  be  so, 
for  there  is  no  loss  in  connection  with  the  impulse  wheel  which 
would  not  vary  in  proportion  to  the  power  of  the  wheel.  Aside 
from  questions  of  design  and  workmanship  the  efficiency  would 
appear  to  be  a  function  of  the  specific  speed.  Too  low  a  value  of 
the  specific  speed  would  mean  a  large  diameter  of  wheel  for  a 
given  power  output  with  a  consequently  large  friction  and 
windage  loss.  Too  high  a  specific  speed  would  mean  that  the 
jet  was  too  large  for  the  wheel  and  buckets  with  a  consequent 
lowering  of  the  hydraulic  efficiency.  The  most  favorable  value 


Size  of  Unit 

FIG.  220. — Effect  of  size  of  turbine  upon  its  efficiency. 

of  N8  is  about  4.0  and  the  best  efficiency  that  is  obtained  is  about 
82.0  per  cent.  This  is  slightly  exceeded  at  times  and  values 
below  it  are  often  obtained. 

With  the  reaction  turbine  the  efficiency  is  a  function  of 
its  size.  This  is  partly  due  to  the  fact  that  the  hydraulic  ef- 
ficiency increases  with  the  size,  but  more  to  the  fact  that  the 
volumetric  efficiency  increases.  With  a  reaction  turbine  there 
is  always  a  certain  amount  of  leakage  between  the  guides  and 
the  runner  so  that  a  portion  of  the  water  escapes  through  the 
clearance  spaces  and  does  not  pass  through  the  wheel.  The 
area  of  these  clearance  rings  would  naturally  be  less  in  proportion 
to  the  area  of  the  wheel  passages  as  the  size  of  the  wheel  in- 
creases. Hence  a  much  larger  per  cent,  of  the  water  is  made 
to  deliver  its  power  to  the  runner.  Such  a  condition  does  not 


TURBINE  LAWS  AND  FACTORS  237 

exist  with  the  impulse  wheel.  This  leads  to  comparative  values 
for  the  two  types  as  shown  in  Fig.  220. 

Another  distinction  between  the  two  types  of  turbines  is  that 
the  reaction  turbine  suffers  certain  hydraulic  losses  on  part 
gate  that  are  lacking  in  the  other.  Hence,  although  in  some 
cases  the  maximum  efficiency  of  a  reaction  turbine  is  greater 
than  that  of  the  impulse  wheel,  the  efficiency  on  a  light  load 
might  not  be  as  good. 

Like  the  impulse  turbine  the  efficiency  of  the  reaction  turbine 
also  depends  upon  the  specific  speed,  being  less  at  either  extreme. 
The  best  efficiencies  are  obtained  with  values  of  N,  ranging  from 
30  to  60.  The  efficiency  of  a  turbine  of  good  design  and  work- 
manship depends  upon  size,  specific  speed,  and  other  factors  to 
such  an  extent  that  definite  values  cannot  be  given,  but  for  fair 
size  units  it  should  range  from  80  to  90  per  cent,  and  occa- 
sionally more.  For  small  wheels,  especially  with  unfavorable 
specific  speeds,  a  value  of  from  60  to  80  per  cent,  is  all  that 
should  be  expected. 

154.  PROBLEMS 

1.  The  turbine,  whose  performance  is  shown  in  Fig.  216,  developed  its 
maximum  efficiency  of  88.0  per  cent,  when  delivering  550  hp.  at  600  r.p.m. 
under  a  head  of  141.8  ft.     The  water  consumed  was  38.8  cu.  ft.  per  sec. 
What  would  be  its  proper  speed  under  a  head  of  283.6  ft.?     What  would 
then  be  the  rate  of  discharge  and  the  horsepower? 

2.  In  Fig.  215  the  turbine  delivered  617  hp.  when  running  at  600  r.p.m. 
under  a  head  of  140.5  ft.,  the  rate  of  discharge  being  44.5  cu.  ft.  per  sec. 
and  the  efficiency  87.0  per  cent.     If  the  speed  is  maintained  at  600  r.p.m. 
when  the  head  is  70.2  ft.,  find  values  of  discharge,  power  delivered,  and 
efficiency.     (Note:     This  can  be  determined  only  by  making  use  of  the 
curves  for  this  particular  turbine.     The  procedure  would  be  to  find  the 
value  of  0  for  the  new  conditions  and  then  take  values  of  q,  hp.,  and  e  from 
the  curves.     These  quantities  would  then  have  to  be  reduced  to  the  proper 
values  for  the  new  head.) 

3.  Compute  the  factprs  <£,  Ki,  and  Ns  for  the  runners  shown  in  Figs. 
152,  157,  167,  170,  and  172.     Compare  these  values  with  each  other. 

4.  It  is  desired  to  develop  6,000  hp.  at  514  r.p.m.  under  a  head  of  625 
ft.     Will  an  impulse  or  reaction  turbine  be  required?     (This  can  be  deter- 
mined by  computing  the  specific  speed.) 

6.  If  only  900  hp.  is  to  be  developed  for  the  conditions  given  in  problem 
(4),  what  type  of  turbine  will  be  required? 

6.  It  is  desired  to  use  a  type  of  turbine  whose  specific  speed  is  30  to 
deliver  100  hp.  under  a  head  of  100  ft.     What  will  be  the  proper  r.p.m. 
for  the  unit? 

7.  An  impulse  wheel  is  to  be  used  for  625  hp.  under  a  head  of  144  ft. 


238  HYDRAULICS 

What  will  be  the  maximum  rotative  speed  at  which  it  can  be  run  without 
material  sacrifice  of  efficiency?  What  will  be  the  approximate  diameter 
of  the  wheel? 

8.  What  would  be  the  minimum  speed  for  a  reaction  turbine  for  the 
conditions  of  problem  (7)?     If  <f>  =  0.60,  what  would  be  the  diameter  of 
the  runner? 

9.  What  would  be  the  maximum  speed  for  a  reaction  turbine  in  problem 
(7)?     Assuming  0  =  0.85,  what  would  be  the  diameter  of  the  runner? 

10.  The  runner  in  the  Cornell  University  turbine  is  27  in.  in  diameter. 
The  wheel  develops  550  hp.  when  running  at  600  r.p.m.  under  a  head  of 
141.8  ft.     What  would  be  the  speed  and  power  of  a  54-in.  runner  of  the 
same  type  under  the  same  head?    Would  the  specific  speed  of  these  two  be 
the  same? 


CHAPTER  XVI 
THE  CENTRIFUGAL  PUMP 

155.  Definition. — Centrifugal   pumps   are   so   called   because 
of  the  fact  that  centrifugal  force  or  the  variation  of  pressure 
due  to  rotation  is  an  important  factor  in  their  operation.1 

In  brief,  the  centrifugal  pump  consists  of  an  impeller  rotating 
within  a  case  as  shown  in  Fig.  221.  Water  enters  the  impeller 
at  the  center,  flows  radially  outward,  and  is  discharged  around 
the  circumference  into  the  case.  During  flow  through  the 
impeller  the  water  has  received  energy  from  the  vanes  resulting 
in  an  increase  both  in  pressure  and  velocity.  Since  a  large  part 
of  the  energy  of  the  water  at  discharge  is  kinetic,  it  is  necessary 
to  conserve  this  kinetic  energy  and  transform  it  into  pressure, 
if  the  pump  is  to  be  efficient. 

As  a  matter  of  convenience  in  illustration,  the  water  is  repre- 
sented as  entering  the  impeller  in  Fig.  221  with  a  positive  pres- 
sure. However,  the  pump  is  usually  set  above  the  level  of  the 
water  from  which  it  draws  its  supply,  in  which  case  the  pressure 
at  this  point  would  be  negative.  Likewise,  the  axis  of  rotation 
need  not  be  vertical  as  shown. 

156.  Classification. — Centrifugal  pumps  are  broadly  divided 
into  two  classes: 

1.  Turbine  pumps. 

2.  Volute  pumps. 

While  there  are  still  other  types  these  two  are  the  most  funda- 
mental. Also,  as  we  shall  see,  these  may  in  turn  be  subdivided 
in  other  ways. 

The  turbine  pump  is  one  in  which  the  impeller  is  surrounded 
by  a  diffuser  containing  stationary  guide  vanes  as  shown  in  Fig. 
222.  These  provide  gradually  enlarging  passages  whose  function 
it  is  to  reduce  the  velocity  of  the  water  leaving  the  impeller  and 
thus  efficiently  transform  velocity  head  into  pressure  head.  The 
casing  surrounding  the  diffuser  may  be  either  circular  and  con- 

1  For  a  more  complete  treatment  of  this  entire  subject,  either  descriptive, 
theoretical,  or  practical,  see  "Centrifugal  Pumps,"  by  R.  L.  Daugherty. 

239 


240 


HYDRAULICS 


centric  with  the  impeller  or  it  may  be  spiral  like  the  cases  of 
some  reaction  turbines. 

The  volute  pump,  shown  in  Fig.  223,  is  one   which   has  no 
diffusion  vanes  but,  instead,  the  casing  is  of  a  spiral  type  so 


FIG.  221. 

proportioned  as  to  produce  an  equal  velocity  of  flow  all  around 
the  circumference  and  also  to  gradually  reduce  the  velocity  of 
the  water  as  it  flows  from  the  impeller  to  the  discharge  pipe. 
Thus  the  energy  transformation  is  accomplished  in  a  slightly 


THE  CENTRIFUGAL  PUMP 


241 


different  way.     This  spiral  is  often  called  a  volute,  whence  the 
name  of  the  pump. 

Occasionally  pumps  have  been  built  with  a  whirlpool  chamber 
as  shown  in  Fig.  221.  This  produces  a  free  spiral  vortex,  the 
nature  of  which  has  been  shown  in  Art.  120. 


FIG.  222. — Turbine  pump. 

157.  Description  of  the  Centrifugal  Pump. — The  centrifugal 
pump  is  similar  to  the  reaction  turbine  both  in  its  construction 
and  in  its  theory.  However,  one  is  not  the  reverse  of  the  other, 
and  their  differences  are  as  striking  as  their  similarities. 


FIG.  223. — Volute  pump. 

The  rotating  part  of  the  pump  which  is  instrumental  in 
delivering  the  water  is  called  the  impeller.  Impellers  may 
receive  water  on  one  side  only  or,  as  in  Fig.  224,  from  both 
sides,  in  which  case  they  are  known  as  double-suction  im- 
pellers. Fig.  225  gives  a  view  of  the  pump  whose  section  is 

16 


242 


HYDRAULICS 


seen  in  Fig.  224.  It  may  be  seen  that  this  impeller  is  relatively 
narrow  as  compared  with  its  diameter,  while  the  opposite  type  is 
shown  in  Fig.  226.  For  the  same  rotative  speed  the  latter  will 
discharge  more  water  than  the  former  but  at  a  lower  head. 

For  high  heads  it  becomes  desirable  to  place  impellers  in  series, 
in  which  case  we  have  the  multi-stage  pump,  such  as  is  shown 
in  Fig.  227.  Multi-stage  pumps  may  be  either  of  the  turbine 
or  the  volute  type.  The  former  may  be  seen  in  Fig.  227  and 
the  latter  in  Fig.  228.  The  addition  of  guide  vanes  so  as  to 
produce  a  turbine  pump  results  in  a  much  more  complex  con- 


Courtesy  of  the  Allis-C halmers  Mfg.  Co. 

FIG.  224. — Double-suction  volute  pump. 

struction,  as  Fig.  229  will  show.  The  water  in  a  multi-stage 
turbine  pump  usually  passes  from  one  impeller  to  the  next 
through  passages  which  are  like  those  shown  in  Fig.  230.  There 
are  other  arrangements  besides  this,  but  they  will  not  be  de- 
scribed here. 

158.  Conditions  of  Service. — Centrifugal  pumps  are  used 
for  lifting  water  a  few  feet  only  or  as  much  as  several  thousand 
feet,  if  necessary.  Several  such  pumps  have  been  built  for  heads 
of  2,000  ft. 

The  capacities  of  centrifugal  pumps  range  from  very  small 
quantities  up  to  as  high  as  300  cu.  ft.  per  sec.  (134,500  G.P.M. 
or  194,000,000  gal.  per  24  hr.).  The  I.  P.  Morris  Co.  has  built 
several  of  the  latter  for  a  head  of  16  ft.  at  77.5  r.p.m. 


THE  CENTRIFUGAL  PUMP 


243 


Courtesy  of  the  Allis-Chalmers  Mfg.  Co. 

FIG.  225. — Double-suction  volute  pump. 


I 


Courtesy  of  Platt  Iron  Wks. 

FIG.  226.— Double-suction  volute  pump. 


244 


HYDRAULICS 


Courtesy  of  Chicago  Pump  Co. 

FIG.  227. — Two-stage  turbine  pump. 


Courtesy  of  Platt  Iron  Wks. 

FIG.  228. — Three-stage  centrifugal  pump  without  diffusion  vanes , 


THE  CENTRIFUGAL  PUMP 


245 


Courtesy  cf  Platt  Iron  Wks. 
FIG.  229. — Two-stage  centrifugal  pump  with  diffusion,  vanes. 


Suction 


FIG.  230. — Worthington  two-stage  turbine  pump. 


246  HYDRAULICS 

The  greatest  power  of  any  centrifugal  pump  is  that  of  a 
pump  installed  by  Sulzer  Bros,  in  Italy.  A  single-stage  pump 
running  at  1,002  r.p.m.  delivers  32,530  G.P.M.  at  a  head  of 
498.6  ft.  with  an  efficiency  of  81.0  per  cent.  The  water  horse- 
power is  3,590  and  the  power  required  to  run  it  is  4,430  hp. 
For  most  pumps  the  power  required  is  less  than  500  hp. 

Rotative  speeds  may  vary  all  the  way  from  30  to  3,000  r.p.m. 
in  ordinary  practice  according  to  circumstances.  The  highest 
speed  ever  employed  was  20,000  r.p.m.  for  a  single-stage  volute 


Courtesy  of  Allis-Chalmers  Mfg.  Co. 

FIG.  231. — 72-in.  centrifugal  pump  for  drainage  at  Memphis,      h  =  15'; 
N  =  100.     Capacity,  194,000.000  gal.  per  day. 

pump  with  an  impeller  2.84  in.  in  diameter.  The  pump  delivered 
250  G.P.M.  against  a  head  of  700  ft.  with  an  efficiency  of  60.0 
per  cent.  The  highest  peripheral  speed  used  was  with  a  single- 
stage  pump  with  an  impeller  3.15  in.  in  diameter.  At  18,000 
r.p.m.  it  delivered  189  G.P.M.  against  a  head  of  863  ft.  and  for 
a  smaller  discharge  it  developed  a  head  of  995  ft. 

Centrifugal  pumps  have  been  built  with  as  many  as  12  stages. 
It  is  customary  to  limit  the  head  per  stage  to  a  value  of  not 
more  than  100  to  200  ft.,  but  this  has  been  greatly  exceeded  in 
several  cases  mentioned  above. 

Water  turbines  are  rated  according  to  the  diameters  of  their 
runners,  but  the  size  of  a  centrifugal  pump  is  usually  designated 


THE  CENTRIFUGAL  PUMP 


247 


by  giving  the  diameter  of  the  discharge  pipe.  The  rated  head 
and  discharge  for  a  centrifugal  pump  are  the  values  for  which 
the  efficiency  is  a  maximum  under  a  given  speed.  This  value 
of  the  rate  of  discharge  is  often  designated  as  the  normal  dis- 
charge. These  values  will  be  different  for  different  speeds. 

159.  Head  Developed. — The  head  developed  by  a  centrifugal 
pump  when  no  flow  occurs  is  called  the  "shut-off  head"  or  the 


Pressure  Chamber, 

^===^ 

FIG.  232. — Crude  centrifugal  pump. 

"head  of  impending  delivery."     Its  value  may  be  found  by 
applying  the  principles  of  Art.  119. 

If  water  in  a  closed  chamber  be  set  in  motion  by  a  paddle- 
wheel  as  in  Fig.  232,  there  will  be  an  increase  in  pressure  from 
the  center  to  the  circumference.  If  the  water  is  assumed  to 
rotate  at  the  same  speed  as  the  impeller,  the  peripheral  velocity 
of  which  is  u2,  it  may  be  seen  from  equation  (117)  that  pz  —  p\  = 
u22/2g,  where  p\  denotes  the  pressure  at  the  center.  If  this  water 
is  in  communication  with  a  pressure  chamber  to  which  a  pie- 


248 


HYDRAULICS 


zometer  tube  is  attached,  as  in  Fig.  232,  water  will  rise  in  the 
latter  to  such  a  height  that 

h  =  u22/2<7  (140) 

If  the  height  of  the  tube  were  less  than  this,  water  would  flow 
out  and  we  should  have  a  crude  centrifugal  pump. 

Actually  there  are  certain  influences  at  work  in  the  real 
pump  which  affect  this  relation  slightly.  Some  of  these  factors 
tend  to  increase  the  head  and  others  to  decrease  it.  The  net 
effect  is  that  for  the  usual  type  of  centrifugal  pump  the  head  of 
impending  delivery  is 

h  =  0.85  to  1.10  uj/2g.  (141) 


130 
120 

110 
100  . 
90 
80 
78 
60 
50 
40 
30 
20 
10 
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Risin 

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Stf>f 

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*X 

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m 

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Spec 

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3 

i 

s 

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\ 

0         10        20        30        40        50        60        70        80        90       100      110      120       130     140 
Percent  of  Normal  Discharge 

FIG.  233. — Head-discharge  characteristics  of  different  pumps. 

But  as  soon  as  flow  occurs  the  above  relation  is  no  longer 
true,  as  may  be  seen  in  Figs.  233  and  234.  When  water  is  being 
delivered,  the  head  may  be  either  greater  or  less  than  the  shut-off 
head,  according  to  the  design  of  the  pump.  We  shall  now 
derive  a  general  relation  between  head,  impeller  speed,  and  rate 
of  discharge  for  all  conditions  of  operation. 


THE  CENTRIFUGAL  PUMP 


249 


As  the  water  in  the  suction  pipe  approaches  the  impeller 
it  may  have  a  rotary  motion  imparted  to  it  before  it  ever  reaches 
the  latter,  due  to  the  viscosity  of  intervening  particles  of  water. 
Hence  we  shall  write  equations  between  points  (2)  and  (s)  in 
Fig.  221,  the  latter  point  being  removed  far  enough  from  the 
impeller  so  that  the  water  has  had  no  rotational  flow  imparted 
to  it.  Let  the  difference  in  elevation  between  these  two  points 
be  denoted  by  z.  Then  from  equation  (118)  we  have 


From  this  it  may  be  seen  that  p2  —  pa  + 
of  equation  140. 


is  equal  to  the  h 


80 


7° 


60 


50 


40 


30 


a 


Efficien 


W.H 


P. 


De  [Laval  Centrifugal  Puinp 
»  1700  R.P.M. 


0       0.2      0.4      0.6      .0.8       1.0      L2      1.4      1.6      1.8       2.0      2.2      .2.4      2.6 
Discharge-Cu.  -Ft.  per  Sec. 

FIG.  234. — Characteristics  of  a  6-in.  pump  at  a  constant  speed. 

Before  proceeding  further  let  us  consider  the  illustration  of  a 
hose  with  a  nozzle  on  the  end.  When  the  nozzle  is  opened  so 
that  water  may  flow,  the  pressure  at  the  base  of  the  nozzle  is 
decreased  below  the  value  obtained  when  it  is  closed  by  an 
amount  equal  to  the  velocity  head  at  that  point  and  to  the  fric- 
tion losses  up  to  that  point.  Now  if  we  should  move  the  hose 
and  the  nozzle  around  while  the  water  is  flowing,  this  pressure 
drop  would  not  be  altered  in  the  least,  for  it  is  a  function  of 
the  velocity  of  flow  within  the  hose,  which  is  the  relative  velocity, 
and  does  not  depend  upon  the  velocity  of  the  water  with  respect 
to  the  earth.  Therefore  at  the  outlet  of  the  impeller  the  pres- 
sure drop  will  be  (1  -f  k")v22/2g,  where  k"  is  a  coefficient  of  loss 
within  the  impeller  passages,  or  from  (s)  to  (2).  But  the 


250  HYDRAULICS 

pressure  at  (s)  likewise  decreases,  when  flow  occurs,  by  an 
amount  equal  to  V82/2g,  and  therefore  the  difference  between  the 
pressures  at  (2)  and  («)  is  decreased  by  (1  +  k")v??/2g  and 
increased  by  V82/2g.  Hence  we  may  now  write 


The  above  equation  enables  us  to  find  the  net  gain  in  pressure 
and  elevation  within  the  impeller. 

But  equation  (142)  does  not  give  the  total  head  developed 
by  the  impeller,  for  we  should  also  have  to  consider  the  change  in 
kinetic  energy,  which  is  proportional  to  (F22  —  Fs2)/2g.  Add- 
ing this  to  the  above  we  have  the  head  developed  between  (s) 
and  (2),  which  is 

uS  -  (1  +  k")  v<?       Vz* 


FIG.  235. — Velocity  diagrams. 

As  may  be  seen  from  the  vector  diagram  of  Fig.  235,  the  value  of 
Vz  is  comparatively  large  and  it  is  necessary  to  convert  this 
kinetic  energy  into  pressure  energy  within  the  diffuser,  volute 
case,  or  whirlpool  chamber.  In  this  way  the  pressure  within 
the  case,  at  (3)  in  Fig.  221,  may  be  greater  than  that  at  the  point 
of  discharge  from  the  impeller.  But  not  all  of  this  kinetic 
energy  can  be  conserved  as  there  are  inevitable  friction  and 
eddy  losses.  We  may  introduce  a  factor  m  to  express  that  por- 


THE  CENTRIFUGAL  PUMP  251 

tion  of  the  velocity  head  that  is  converted,  so  that  the  total 
head  developed  by  the  pump,  including  impeller  and  case,  is 

m'-d  +  ngi1 ,    TV 

~2iT  +mV 

It  must  be  noted  that  the  quantity  m  is  a  variable.  When 
the  discharge  from  the  impeller  is  such  that  the  angle  A2  (see 
Fig.  235)  agrees  with  the  angle  of  the  diffusion  vanes  of  a  turbine 
pump,  or  the  velocity  Y2  is  the  proper  value  for  a  volute  pump, 
the  maximum  proportion  of  the  velocity  head  will  be  saved. 
For  larger  or  smaller  discharges  than  this  there  will  be  additional 
losses  attending  this  conversion.  For  a  turbine  pump  the 
maximum  value  of  m  is  about  0.75  and  for  a  volute  pump  it  is 
somewhat  less. 

With  a  centrifugal  pump  the  impeller  areas  are  fixed  and 
constant  in  value  and  hence  it  is  convenient  to  express  the 
rate  of  discharge  as 

q  =  M.  (144) 


^1 

(c) 

^"  u» 

'a 
FIG.  236. — Stream  lines  for  three  different  rates  of  discharge. 

Now  referring  to  the  vector  diagrams  shown  in  Figs.  235  and 
236  it  may  be  noted  that  as  the  rate  of  discharge  varies  the 
values  of  Vz  and  A  2  change.  It  may  be  seen  that  as  q  ap- 
proaches zero,  Vz  and  A2  approach  zero,  while  Vz  approaches 
HZ.  Hence  for  an  infinitesimal  discharge  the  value  of  Vz  may  be 
regarded  as  equal  to  Hz,  while  the  velocity  of  the  water  in  the 
case  surrounding  the  impeller  is  practically  zero.  Therefore,  a 
particle  of  water  leaving  the  impeller  with  a  high  velocity  enters 


252  HYDRAULICS 

a  body  of  water  at  rest  and  loses  all  of  its  kinetic  energy.  Thus, 
as  the  rate  of  discharge  approaches  zero,  the  factor  m  approaches 
zero.  Hence  it  may  be  seen  that  when  we  have  zero  discharge 
the  value  of  h  in  equation  (143)  reduces  to  that  given  by  equa- 
tion (140). 

An  inspection  of  equation  (143)  serves  to  explain  the  rising 
or  falling  characteristics  of  Fig.  233.  If  the  increase  of  pressure 
due  to  the  conversion  of  the  velocity  head  of  discharge  is  more 
than  enough  to  offset  the  decrease  due  to  the  velocity  and  the 
losses  within  the  impeller,  we  have  a  rising  characteristic.  If 
they  are  about  equal  we  have  a  flat  characteristic,  and  if 
the  quantity  mVz2  is  less  than  (1  -f-  kff)v22  we  have  a  falling 
characteristic. 

Because  of  the  difficulty  of  efficiently  transforming  velocity 
head  into  pressure  head  it  is  desirable  to  keep  F2  as  small  as 
possible.  It  may  be  seen  that,  for  a  given  value  of  vz,  the  smaller 
the  angle  a2  the  less  will  be  the  magnitude  of  F2.  Therefore,  in 
almost  all  centrifugal  pumps  the  value  of  a2  is  from  20°  to  30°, 
though  occasionally  this  angle  is  as  small  as  10°  or  as  large  as 
80°.  It  is  rarely  made  larger  than  90°  because  of  the  inefficiency 
of  such  designs.1 

160.  Measurement  of  Head. — We  may  compute  the  head 
which  a  pump  is  required  to  work  against  by  equations  (72)  or 
(74).  The  head  which  the  pump  can  develop  may  be  estimated 
by  equation  (143),  but  when  we  desire  to  measure  the  head  which 
the  pump  actually  does  develop  we  do  so  by  taking  certain  read- 
ings on  the  discharge  and  suction  sides  of  the  pump.  Thus  in 
Fig.  237  the  difference  between  the  energy  which  the  water  has 
as  it  enters  the  pump  at  (s)  and  that  with  which  it  leaves  at  (d) 
is  due  solely  to  the  pump.  Hence  we  may  write 

h  =  Hd  -  H8. 

But  Hd  =  pd  +  zd  +  Vd*/2g,  and  H8  =  pa  +  zs  +  Va2/2g. 
Therefore  it  follows  that 

h  =  (pd  -  ps)  +  (zd  -  2.)  +  (TV  -  V.2)/20.         (145) 

1  In  the  turbine  theory  the  angle  a  has  been  defined  as  the  angle  between 
v  and  u.  This  is  satisfactory  for  that  purpose,  but  with  ordinary  centrifugal 
pumps  this  angle  is  always  greater  than  90°.  Hence  it  is  much  more 
convenient  to  define  it  here  as  the  angle  between  v  and  —  u,  as  may  be  seen 
in  Fig.  235. 


THE  CENTRIFUGAL  PUMP 


253 


As  a  usual  thing  the  water  enters  the  pump  under  a  pressure 
less  than  that  of  the  atmosphere,  in  which  case  the  value  of  ps 
will  be  negative.  If  the  suction  and  discharge  pipes  at  the 
points  where  the  gages  are  attached  are  of  the  same  diameter 
the  velocity  heads  will  cancel,  in  which  event  the  value  of  h 


FIG.  237. — Head  developed  by  pump. 


pitot  Tube 


FIG.  238. — Measurement  of  head. 

will  be  the  difference  in  the  levels  of  the  surfaces  of  the  two 
water  columns  shown  in  Fig.  237. 

In  testing  the  pump  the  gages  might  be  connected  as  shown 
in  Fig.  238.  It  is  not  really  necessary  to  reduce  the  gage  read- 
ings to  the  pressures  which  would  be  found  at  the  center  line 


.  254  HYDRAULICS 

of  the  pipe.  If  the  gage  readings  are  used  direct  in  equation 
(145)  and  the  value  of  y  represents  (zd  —  zs),  it  may  be  shown 
that  the  result  is  the  same.1 

161.  Head  Imparted  by  Impeller.  —  The  amount  of  energy 
delivered  to  the  water  by  the  impeller  is  greater  than  that 
actually  delivered  in  the  water,  the  difference  being  due  to 
hydraulic  friction  losses  within  the  pump.  If  the  head  actually 
developed  by  the  pump  is  represented  by  h,  the  head  imparted 
to  the  water  by  the  impeller  is 

h"  =  h  +  h'    or    h"  =  h/eh. 

In  an  ideal  pump  without  hydraulic  losses  of  any  kind  these  two 
quantities  would  be  equal,  but  in  any  real  pump  they  represent 
two  entirely  different  things.  For  a  given  pump  under  different 
conditions  of  operation,  h"  and  h  neither  differ  from  each  other 
by  a  constant  amount  nor  is  one  a  constant  proportion  of  the 
other.  Hence  the  curves  representing  actual  values  of  both  h" 
and  h  not  only  do  not  coincide  but  they  are  not  even  of  the  same 
shape.  This  may  be  seen  in  Fig.  239  in  which  the  curve  "Actual 
Head  Input  —  h""  has  been  determined  with  a  reasonable  degree 
of  accuracy  from  test  data  by  the  author.  It  may  be  seen 
that  for  rates  of  discharge  from  0  to  0.6  cu.  ft.  per  sec.  the  value 
of  h  increases  while  that  of  h"  decreases.  In  some  other  cases 
the  difference  is  more  marked  than  is  here  shown. 

An  expression  for  h"  may  be  derived  by  using  the  value  of 
T  given   in   equation    (102).     Since    Wh"  =  Tw,    h"  = 
The  value  of  sz  is  uz  —  v%  cos  a2,  hence 

-  V2  cos  a2) 


a 


^ 


1  The  question  is  often  raised  as  to  why  it  is  necessary  to  deduct  Vf2/2g 
in  determining  the  head,  since  the  pump  has  imparted  that  velocity  to  the 
water.  The  first  answer  is  that  equation  (145)  is  the  result  of  a  direct 
application  of  the  principles  of  energy,  but  the  explanation  of  the  matter  is 
that  we  also  have  included  ps,  whose  value  is  a  function  of  V,.  Suppose, 
for  example,  that  the  suction  pipe  were  so  large  that  the  velocity  in  it 
were  negligible.  Then  the  measured  value  of  ps  would  give  a  higher  pres- 
sure than  when  the  suction  pipe  is  smaller  and,  disregarding  losses,  the 
values  of  ps  in  the  two  cases  would  differ  by  V^/lg.  If  we  are  to  omit  the 
velocity  head  at  (s)  we  should  omit  the  pressure  reading  also.  We  might 
then  obtain  the  total  head  by  adding  to  the  "discharge  head"  the  value  of 
zs  +  suction  pipe  losses.  But  we  should  have  to  compute  the  latter  and 
it  may  be  shown  that  they  are  determined  experimentally  when  ps  is  meas- 
ured and  the  velocity  head  Vs2/2g  also  employed. 


THE  CENTRIFUGAL  PUMP 


255 


This  could  also  be  obtained  from  equation  (143)  by  eliminating 
the  hydraulic  losses.  This  would  require  values  of  k"  =  0  and 
m  =  1.0.  The  next  step  would  be  to  solve  the  vector  triangle 
for  V-i  in  terms  of  u%,  v2)  and  a2.  The  result  would  agree  with 
equation  (146). 


140 


0       0.2      0,4      0.6      0.8      1.0     1.2      1.4      1.6      1.8      2.0      2.2     2.4      2.6 
Discharge-Cu.  ft.  per  Second. 

FIG.  239. — Analysis  of  centrifugal  pump  at  a  constant  speed. 


It  may  be  seen  that  for  a  constant  value  of  impeller  speed  the 
value  of  h"  given  by  equation  (146)  will  increase  with  q  (and  ^2) 
for  values  of  a2  greater  than  90°,  will  be  independent  of  q  for 
a2  =  90°,  and  will  decrease  for  values  of  a2  less  than  90°.  It  is 
sometimes  argued  from  this  that  rising  or  falling  characteristics 
are  obtained  by  a  suitable  choice  of  a2,  but  that  is  due  to  confusing 


256  HYDRAULICS 

h"  and  h.  The  value  of  a2  does  have  some  effect  upon  this,  but 
it  alone  does  not  determine  the  matter.  The  author  has  found  a 
decidedly  rising  characteristic  in  a  pump  he  has  tested  with  an 
angle  of  a?  =  26°.  And  tests  of  other  pumps  with  a2  =  90° 
have  shown  steep  falling  characteristics.  The  real  explanation 
may  be  seen  only  in  equation  (143). 

The  hydraulic  efficiency  is  the  ratio  h/h".  For  the  same 
reasons  as  are  given  in  Art.  142,  it  is  difficult  to  calculate  true 
values  of  In!'  and  thus  the  true  hydraulic  efficiency  can  be  deter- 
mined only  by  test.  The  latter  gives  directly  the  value  of  the 
total  efficiency  only  and  it  is  necessary  to  allow  for  other  losses 
or  determine  them  by  special  methods  in  order  to  get  the  actual 
hydraulic  efficiency.  Applying  equation  (146)  and  using  the 
actual  impeller  dimensions  the  computed  values  of  h"  may  be 
found  to  lie  on  a  straight  line,  such  as  is  labelled  "Ideal  Head 
Input1'  in  Fig.  239.  Thus  the  ratio  between  the  actual  h  and 
the  h"  computed  in  the  ordinary  manner  is  much  less  than  the 
hydraulic  efficiency  in  all  cases.  It  is  very  often  less  than  the 
gross,  efficiency  thus  proving  that  it  is  not  a  true  value.  But 
it  is  still  useful  for  some  purposes  of  design  and  is  called  "mano- 
metric  coefficient.''  The  value  of  this  ratio  is  usually  between 
0.55  and  0.65. 

162.  Centrifugal-pump  Factors. — Just  as  in  the  case  of  turbines, 
it  is  found  that  to  obtain  the  best  efficiency  with  a  given 
centrifugal  pump  there  must  be  a  certain  relation  between  head, 
speed,  and  discharge.  Also  the  equations  show  that  these 
three  quantities  are  mutually  interrelated.  Hence  it  may  be 
seen  from  equation  (143)  that  for  a  velocity  diagram  of  the  same 
shape  to  be  formed  it  is  necessary  that  w2,  w2,  and  72  vary  as 
Vh.  Hence  we  shall  find  it  convenient  to  write1 

u2  =  <!>V2gh  (147) 

v2  =  c\/2gh  (148) 

We  find  that  a  certain  value  of  0  is  required  to  obtain 
the  maximum  efficiency  just  as  with  the  turbine.  And  a 
definite  value  of  c  is  associated  with  every  value  of  0  as  may 

1  For  the  pump  0  has  the  same  meaning  as  in  the  inward  flow  reaction 
turbine  since  it  gives  the  peripheral  speed  in  both  cases.  But  c  has  a 
different  meaning,  since  we  find  it  more  convenient  to  deal  with  y2  rather 
than  with  W 


THE  CENTRIFUGAL  PUMP  257 

be  seen  from  equation   (143).     For  ordinary  types  of  pumps 
we  find  the  following  values  of  these  factors: 

For  shut-off  <£  =  0.95  to  1.09 

For  normal  discharge   <f>e=  0.90  to  1.30 
For  normal  discharge  ce  =  0.10  to  0.30 

The  value  of  fa  will  depend  upon  the  design  of  the  pump. 
Thus  the  smaller  the  angle  a2  and  the  fewer  the  number  of 
impeller  vanes,  the  larger  the  value  of  </>e. 

Just  as  in  Art.  150,  it  may  be  shown  that 

(137) 


163.  Specific  Speed.  —  The  specific  speed  factor  for  turbines 
involves  the  developed  horsepower,  since  that  is  the  quantity 
with  which  we  are  concerned.  But  with  centrifugal  pumps  we 
are  primarily  interested  in  their  capacity  and  it  will  be  more 
useful  if  we  derive  a  similar  expression  giving  Na  in  terms  of 
discharge.  Since  power  and  discharge  are  really  proportional 
to  each  other  it  may  be  seen  that  we  are  merely  expressing 
the  specific  speed  for  the  pump  in  terms  of  different  units. 

Proceeding  just  as  in  Art.  151,  except  that  we  use  equation 
(138)  direct,  we  obtain  for  the  centrifugal  pump 


(149) 

The  capacity  of  a  centrifugal  pump  is  generally  expressed  in 
gallons  per  minute  rather  than  in  cubic  feet  per  second.  Thus 
the  expression  will  probably  be  the  handiest  in  the  above  form. 
(1  cu.  ft.  =  7.48  U.  S.  gal.). 

For  an  impeller,  either  single-suction  or  double-suction, 
values  of  specific  speed  may  be  found  between  the  following 
limits:  ' 

N8  =  500  to  8,000. 

For  special  constructions  even  higher  values  may  be  attained. 
It  must  be  noted  that  these  apply  only  to  single  stages.  For  a 
multi-stage  pump  it  is  necessary  to  divide  the  total  head  by  the 
number  of  stages  to  obtain  the  proper  value  of  h  for  use  in  the 
equations  in  this  chapter. 


*Note  that  h**  =  h  +  hy*  =  h  -r-  \/\/.     Some  values   of  hy*  will  be 
found  on  page  263. 


17 


258  HYDRAULICS 

Values  of  specific  speeds  are  obtained  from  tests  of  actual 
pumps  and  they  may  then  be  applied  to  other  pumps  of 
the  same  type.  For  N8  is  an  index  of  the  type  of  pump  just  as 
it  is  in  the  case  of  the  turbine.  Its  great  value  is  that  it  enables 
us  to  determine  the  combinations  of  speed,  capacity,  and  head 
per  stage  that  are  possible  or  desirable.  And  if  we  desire  to 
employ  a  certain  type  of  pump  with  a  definite  value  of  Ns,' 
we  may  then  find  the  combinations,  of  these  factors  that  are 
required. 

164.  Operation  at  Different  Speeds. — In  this,  chapter  we 
have  shown  characteristics  of  centrifugal  pumps  operating 
under  variable  heads  at  constant  speeds.  We  may  now  desire 
to  know  how  the  pump  is  affected  by  a  change  in  speed.  This  is 
shown  by  equations  (147)  and  (148).  To  obtain  similar 
conditions  of  operation  it  is  necessary  that  the  values  of  0 
and  c  be  maintained  constant.  If  they  are,  it  may  be  seen  that 
both  the  speed  and  discharge  of  the  pump  will  vary  as  the  square 
root  of  the  head.  But  if  $  and  c  are  not  constant  then  we  have 
no  simple  index  to  the  variation  of  the  e  quantities.  We  can  then 
only  resort  to  some  secDnd  degree  equation  of  the  form  shown  in 
Art.  159.  Hence  if  the  head  is  varied  due  to  a  change  in  speed 
it  must  be  understood  that  the  rate  of  discharge  varies  also  if 
the  following  simple  ratios  are  to  apply. 

From  equation  (147)  it  may  be  seen  that 

»  -  £  If  (150 

Which  shows  that  if  0  remains  constant,  the  head  developed 
varies  as  the  square  of  the  pump  speed.  From  equation  (148) 
we  may  obtain,  after  substituting  the  value  of  h  given  by  equa- 
tion (150), 

V2   =   %2  *      (151) 

9 

Hence  it  follows  that  if  <j>  remains  constant,  c  will  also  remain 
constant  and  the  rate  of  discharge  must  vary  directly  as  the 
speed.  Since  power  is  a  function  of  the  product  of  h  and  q  it 
may  be  seen  that  it  will  vary  as  the  cube  of  the  speed.  Just  as 
in  the  case  of  the  turbine,  the  hydraulic  efficiency  of  a  centrifugal 
pump  is  independent  of  the  speed,  within  reasonable  limits,  as 
long  as  </>  is  constant.  But  the  maximum  gross  efficiency  of  a 
given  pump  will  increase  slightly  as  higher  speeds  are  attained. 


THE  CENTRIFUGAL  PUMP 


259 


165.  Factors  Affecting  Efficiency. — The  considerations  of 
Art.  153  apply  here  also.  The  most  important  factor  in  deter- 
mining the  efficiency  of  a  centrifugal  pump  is  its  capacity,  as 
may  be  seen  in  Figs.  240  and  241.  A  pump  of  small  capacity 
will  have  a  low  volumetric  efficiency  because  of  the  relatively 
large  per  cent,  of  the  water  which  will  leak  back  into  the 


0         1,000       2,000       3,000       4,000       5,000       6,000       7,000       8,000       9,000      10,000 
Turbine  Pumps,  Discharge  in  Gal/per  Min. 

FIG.  240. — Efficiency  as  a  function  of  capacity. 

suction  side  through  the  clearance  rings.  Also  the  disk  fric- 
tion of  such  a  pump  is  a  greater  percentage  of  the  total  power 
expended. 

It  may  be  shown  that  the  head  per  stage  has  only  a  slight 
effect  upon  the  efficiency  of  the  pump,  providing  the  design  is 
carefully  made. 


IDOj  

. 

• 

• 

• 

i4x 

j-  en\  1. 

>* 

: 

. 

o  W  / 

gJlL 

IT 

«E  40  1 

30  1 

201— 

2.000   4,000   6,000   8,000   10.000   12,000   14,000  16,000  18,000  20,000 

Volute  Pumps,  Discharge  in  Gal.  per  Min.  * 

FIG.  241. — Efficiency  as  a  function  of  capacity. 

For  a  given  capacity,  however,  the  efficiency  will  be  found  to 
differ  with  different  pumps,  due  not  only  to  variations  in  work- 
manship and  construction  but  also  to  the  other  factors  such  as 
speed  and  head.  Since  these  are  all  involved  in  the  specific 
speed,  it  would  seem  reasonable  that  efficiency  may  be  expressed 


260 


HYDRAULICS 


as  a  function  of  the  latter.  Figs.  242  and  243  show  the  relation 
between  efficiency  and  specific  speed  for  a  large  number  of 
turbine  and  volute  pumps.  But  it  should  be  borne  in  mind 
that  for  any  given  specific  speed  the  larger  the  capacity  the  higher 
the  efficiency.  Hence  we  can  have  no  single  curve  that  will 
enable  us  to  select  definite  values  for  any  case. 


90 


X^ 


0     200   400    600  800  1000  1200  1400  1600 1800  2000  2200  2400  2600  280030003200  3400  3600 
Turbine  Pumps     Specific  Speed  Ns^  NyG.P^. 

FIG.  242. — Efficiency  as  a  function  of  specific  speed. 


Efficiency  Percent 

g  S  §  §  3  S  3 

Jl 

^-•• 

:  — 

'^ 

•,     — 

—  i. 

T-] 

••     ,• 

-^—  . 

-k4 

X 

ff  • 

• 

-*-? 

—  .  ~. 

•  — 

—  •  — 

/ 

/ 

/• 

X 

,- 

—  • 

— 



—  —  .  . 



"7- 

—  . 

-^^_ 

— 

/ 

/ 

/ 

f 

/ 

/ 

/ 

0  400  800  1200  1600  2000  2100  2800  3200  3600  4000  4400  4800  5200  5600  6000  6400  €800  7200  7600  i 

Volute  Pumps,     Specific  Speed,      N,  =  N<v^^Mf 

FIG.  243. — Efficiency  as  a  function  of  specific  speed. 


166.  PROBLEMS 

1.  The  curves  of  Fig.  239  are  for  a  single-stage  pump  in  which  D  =9.12 
in.,  /2  =  0.0706  sq.  ft.,  a2  =  27°.     At  1,700  r.p.m.  when  q  =  1.315  cu.  ft. 
per  second,  h  =  55.7  ft.     If  it  be  assumed  that  ra  =  0.50,  find  the  value 
of  k". 

2.  A  two-stage  turbine  pump  running  at  1,700  r.p.m.  delivered  0.429 
cu.  -ft.  per  sec.  at  a  head  of  225  ft.     The  essential  dimensions  were:  D  = 
12  in.,  /2  =  0.0244  sq.  ft.,  a2  =  26°.     If  it  be  assumed  that  m  =  0.70, 
compute  the  value  of  k". 

3.  The  curves  of  Fig.  234  were  obtained  from  the  test  of  a  single-stage 
pump  having  the  following  dimensions:  D  =9.12  in.,  /2  =  0.0706  sq.  ft. 
Find  the  values  of<&  and  ce. 

4.  If  it  be  assumed  in  Fig.  233  that  the  value  of  <£  for  shut-off  is  1.0,  what 
is  the  value  of  0  for  the  maximum  lift  of  the  pump  with  the  rising  char- 
acteristic?    What  is  the  value  of  <£  for  maximum  efficiency  in  each  case? 


THE  CENTRIFUGAL  PUMP  261 

6.  The  diameter  of  a  pump  impeller  is  10  in.     The  speed  is  to  be  1,200 
r.p.m.     If  </>  =  1.20,  what  is  the  value  of  hi 

6.  Compute  the  value  of  the  specific  speed  for  the  pump  shown  in  Fig. 
231. 

7.  Compute  the  value  of  the  specific  speed  for  the  pump  whose  dimensions 
are  given  in  problem  (2). 

8.  What  would  be  the  capacity,  head,  and  power  of  the  pump  whose 
performance  is  shown  in  Fig.  234,  if  it  were  run  at  a  speed  of  1,000  r.p.m.? 

9.  What  speed  would  be  necessary  to  double  the  capacity  of  the  pump 
whose  curves  are  shown  in  Fig.  234?     What  speed  would  be  required  to 
double  its  lift? 

10.  If  the  speed  of  the  pump  of  Fig.  234  were  doubled,  what  would  be 
the  head  for  a  discharge  of  2.4  cu.  ft.  per  sec.?     What  would  be  the  ef- 
ficiency for  this  rate  of  discharge  at  the  higher  speed? 

11.  It  is  desired  to  deliver  1,600  G.P.M.  at  a  head  of  900  ft.  with  a  single- 
stage  pump.     What  would  be  the  minimum  rotative  speed  that  could  be 
used? 

12.  If  a  speed  of  600  r.p.m.  is  desired  in  problem  (11),  how  many  stages 
must  the  pump  have  at  least? 

13.  It  is  desired  to  use  a  type  of  pump  whose  specific  speed  is  2,000 
under  a  Head  of  16  ft.     If  the  speed  is  to  be  1,800  r.p.m.,  what  will  be  the 
capacity? 

14.  Compute  the  specific  speeds  of  the  pumps  for  which  data  are  given 
in  Art.  158. 


262 


HYDRAULICS 


APPENDIX— TABLES 

TABLE  9. — AREAS  OF  CIRCLES 


Diameter                                     Area 

Dian 

Inches 

ieter 

Area 

Inches 

Feet 

Square 
inches 

Square 
feet 

Feet 

Square 
inches 

Square 
feet 

H 

0  .  0021 

0.0491 

0.00034 

30 

2.500 

706.9 

4.90 

H 

0.0042 

0  .  1963 

0.00136 

32 

2.667 

804.3 

5.58 

% 

0.0062 

0.4417 

0  .  00306 

34 

2.830 

907.9 

6.30 

i 

0.083 

0.7854 

0.00545 

36 

3.000 

1,018.0 

7.07 

1H 

0.104 

1.227 

0.00853 

38 

3.17 

1,134.0 

7.88 

1M 

0.125 

1.767 

0.0123 

40 

3.44 

,257.0 

8.72 

134 

0.146 

2.405 

0.0167 

42 

3.50 

,385.0 

9.62 

2 

0.167 

3.142 

0.0218 

44 

3.67 

,521.0 

10.57 

2^ 

0.208 

4.909 

0.0341 

46 

3.83 

,662.0 

11.53 

3 

0.250 

7.069 

0.0492 

48 

4.00 

,810.0 

12.07 

3^ 

0.292 

9.621 

0.0668 

50 

4.17 

,964.0 

13.63 

4 

0.333 

12.566 

0.0872 

52 

4.33 

2,124.0 

14.75 

4>£ 

0.375 

15.909 

0.1105 

54 

4.50 

2,290.0 

15.90 

5 

0.417 

19.635 

0.1362 

56 

4.67 

2,463.0 

17.10 

6 

0.500 

28.27 

0.196 

58 

4.83 

2,642.0 

18.35 

7 

0.583 

38.48 

0.267 

60 

5.00 

2,827.0 

19.62 

8 

0.667 

50.26 

0.349 

62 

5.17 

3,019.0 

20.93 

9 

0.750 

63.62 

0.442 

64 

5.33 

3,217.0 

22.3 

10 

0.833 

78.54 

0.545 

66 

5.50 

3,421.0 

23.8 

12 

1.000 

113.1 

0.785 

68 

5.67 

3,632.0 

25.2 

14 

1.167 

153.9 

1.068 

70 

5.83 

3,848.0 

26.7 

16 

1.333 

201.1 

1.395 

72 

6.00 

4,072.0 

28.3 

18 

1  .  500 

254.5 

1.765 

76 

6.33 

4,536.0 

31.4 

20 

1.667 

314.2 

2.18 

80 

6.67 

5,027.0 

34.9 

22 

1.833 

380.1 

2.64 

90 

7.50 

6,362.0 

44.2 

24 

2.000 

452.4 

3.14 

100 

8.33 

7,854.0 

54.5 

26 

2.164 

530.9 

3-68 

110 

9.17 

9,503.0  66.0 

28 

2.332 

615.8 

4.27 

120 

10.0 

11,310.0 

78.5 

THE  CENTRIFUGAL  PUMP 


263 


TABLE  10. — STANDARD  WROUGHT-IRON  PIPE  SIZES 


Diameter 

Internal  area 

Diameter 

Internal  area 

Nominal, 
inches 

Actual 
internal* 
inches 

Square 
inches 

Square 
feet 

Nominal, 
inches 

Actual 
internal, 
inches 

Square 
inches 

Square 
feet 

1A 

0.27 

0.0573 

0.0004 

3^2 

3.548 

9.887 

0.0687 

H 

0.364 

0.1041 

0.0007 

4 

4.026 

12.73 

0.0884 

H 

0.494 

0.1917    0.0013 

i« 

4.508 

15.96 

0.1108 

M 

0.623 

0.3048 

0.0021 

5 

5.045 

19.99 

0.1388 

% 

0.824 

0.5333 

0.0037 

6 

6.065 

28.89 

0.2006 

l 

1.048 

0.8626 

0.0060 

7 

7.023 

38.74 

0.2690 

IK 

1.380 

1.496 

0.0104 

8 

7.982 

50.04 

0.3474 

11A 

1.611 

2.038 

0.0141 

9 

8.937 

62.73 

0.4356 

2 

2.067 

3.356 

0.0233 

10 

10.019 

78.84 

0.5474 

2U 

2.468 

4.784 

0.0332 

11 

11.000 

95.03 

0.6600 

3 

3.067 

7.388 

0.0513 

12 

12.000 

113.1 

0.7854 

TABLE  11. — VALUES  OF 


m 

•* 

m 

m* 

m 

ptt 

m 

m^ 

0.2 

0.342 

0.2 

1.69 

4.0 

2.52 

8.0 

4.00 

0.4 

0.543 

0.4 

1.79 

0.2 

2.60 

0.5 

4.17 

0.6 

0.712 

0.6 

1.89 

0.5 

2.73 

9.0 

4.33 

0.8 

0.863 

0.8 

1.98 

5.0 

2.92 

10.0 

4.63 

1.0 

1.000 

3.0 

2.08 

0.5 

3.12 

11.0 

4.93 

0.2 

1.13 

0.2 

2.17 

6.0 

3.29 

12.0 

5.22 

0.4 

1.25 

0.4 

2.26 

0.5 

3.48 

13.0 

5.52 

0.6 

1.37 

0.6 

2.35 

7.0 

3.66 

14.0 

5.80 

0.8 

1.48 

0.8 

2.44 

0.5 

3.83 

is'o 

6.10 

2.0 

1.58 

TABLE  12. — VALUES  OF 


h 

& 

h 

fc3/4 

h  • 

*N 

h 

** 

10 

5.62 

25 

11.18 

70 

24.20 

140 

40.6 

11 

6.03 

30 

12.82 

80 

26.77 

150 

42.8 

12 

6.45 

35 

14.38 

90 

29.33 

170 

47.1 

13 

6.85 

40 

15.90 

100 

31.6 

200 

53.2 

14 

7.24 

45 

17.38 

110 

33.9 

230 

59.0 

16 

8.00 

50 

'  18.80 

120 

36.2 

260 

64.8 

18 

8.73 

60 

21.25 

130 

38.5 

300 

72.0 

20 

9.45 

264  HYDRAULICS 

FUNDAMENTAL  TRIGONOMETRY 
In  a  right  angle  triangle,  such  as  Fig.  244: 

sin  A  =  a/c  sec  A  =  c/b 


cos  A  =  b/c 
tan  A  =  a/b 


esc  A  =  c/a 
cot  A  =  b/a 


Any  function  of  A  is  the  same  numerically  as  the  co-function 
of  any  combination  of  A  with  an  odd  multiple  of  90°.     Thus: 

sin  A  =  cos  (90°  ±  A)  =  cos  (270°  ±  A). 

Any  function  of  A  is  the  same  numerically  as  the  function 
of  any  combination  of  A  with  an  even  multiple  of  90°.     Thus: 

sin  A  =  sin  (180°  ±  A). 

The  sign  of  the  function  depends  in  any  case  upon  the  quadrant 
in  which  the  angle  itself  lies. 

B 


FIG.  245. 


For  the  solution  of  an  oblique  triangle,  such  as  that  shown  in 
Fig.  245,  we  have 

sin  A  _  sin  B  _  sin  C 

a  b  c 

a*  =  b2  +  c2  -  2bc  cos  A. 

This  is  as  much  as  is  required  for  the  solution  of  the  vector 
triangles  that  will  be  encountered  with  turbines  and  centrifugal 
pumps. 


INDEX 


Absolute  path,  153,  157,  240, 

pressure,  12 

velocity,  152 
Air  chamber,  151 
Air  in  pipes,  105 
Angular  momentum,  156 
Atmospheric  pressure,  11 
Automatic  crest,  39 

B 

Barometer,  11 
Bazin,  87,  132 
Bends,  pressure  at,  144 
Bernoulli's  theorem,  50 
Buckets,  166 
Buoyancy,  center  of,  31 
force  of,  30 


Case,  189 

Center  of  pressure,  20 
Centrifugal  action,  161 
Characteristics,  pump,  218 
Chezy's  formula,  107 
Cippoletti  weir,  89 
Coefficient  of  contraction,  63 

of  discharge,  64 

of  velocity,  62 
Compressibility  of  water,  2 
Continuity,  equation  of,  48 
Contracted  weir,  84 
Critical  velocity,  46 
Current  meter,  93 


Differential  manometer,  16 
Diffuser,  239 
Discharge  loss,  100 
Discharge,  measurement  of,  92 

rate  of,  48 
Draft  tube,  185 
Dynamic  force,  141 


E 


Effective  head,  52 

Efficiency,  factors  affecting,  236,  259 

as  functions  of  size,  236,  259 

of  pipe  line,  113 

pump,  161    ' 

turbine,  161 
Energy,  54 
Entrance  losses,  97 


Flashboards,  39 
Flow,  line,  199 

non-uniform,  135 

steady,  47 

uniform,  127 

unsteady,  148 
Fluid,  2 

Force  exerted,  104,  143,  215 
Forced  vortex,  161 
Forebay,  201 
Francis  turbine,  177 
Francis  weir  formula,  86 
Free  surface,  2 
Free  vortex,  163 
Friction  factors,  110. 


Dams,  32 
Density  of  water,  3 


Gage  height,  138 
Gage  pressure,  12 
Gates,  182 


265 


266 


INDEX 


Governing,  172,  185. 
Gradient,  hydraulic,  104 
Guide  vanes,  178,  183 

H 

Hazen-Williams,  108 

Head,  developed  by  pump,  114,  251 

for  impulse  wheel,  219 

for  reaction  turbine,  229 

losses  of,  96 

meaning  of,  52 

measurement  of  pump,  252 

shut-off,  247 
Hook  gage,  80 
Hydraulic  gradient,  104,  128 

mean  depth,  97 

radius,  97 

slope,  107 


Ideal  velocity,  65 

Impeller,  241 

Impending  delivery,  247 

Impulse  of  jet,  153 

Impulse  turbine,  155.,  165,  209 

Intensity  of  pressure,  7 


Jet,  coefficients  of,  63 
definition  of,  62 
force  of,  153 
power  of,  55 

K 

Kinetic  energy,  true,  57 
Kutter's  formula,  129 


Liquid,  definition,  2 
Losses  of  head,  96 

M 
Manning's  formula,  131 


Manometric  coefficient,  256 
Measurement  of  head,  14,  252 
Metacenter,  31 

N 

Needle  nozzle,  170 
Non-uniform  flow,  135 
Nozzles,  71,  170 

efficiency  of,  73 

loss  in,  100 


Open  channel,  125,  132 
Orifice,  65 

coefficients,  69 


Path  of  water,  153,  157 
Pelton  wheel,  165 
Penstock,  201 
Piezometer,  14 
Pipe  efficiency,  113 

friction,  97,  108 
Pitot  tube,  91,  145 
Power,  159,  217 

delivered  by  pipe,  113 

meaning  of  head,  55 

plants,  198 
Press,  hydraulic,  17 
Pressure,  absolute,  17 

on  area,  19 

gage,  17 

intensity,  7 

negative,  13 

wave,  148 
Pump,  characteristics,  218 

service,  242 

turbine,  239 

volute,  240 

R 

Radius,  hydraulic,  97 

Rate  of  discharge,  48 

Rating  curve,  138 

Reaction  of  jet,-  155 

Reaction  turbine,  155,  177,  221 


Relative  velocity,  152 
Runner,  179 


Settings,  turbine,  188 
Shut-off  head,  247 
Siphon,  123 
Slope,  hydraulic,  107 
Specific  speed,  234,  257 

uses  of,  235 
Speed,  218,  226,  258 
Standard  orifice,  67 
Steady  flow,  47 
Stream  gaging,  136 
Suppressed  weir,  81 
Surge,  148 
Surge  chamber,  151 


Tangential  wheel,  165 
Torque,  157,  158,  224 
Trapezoidal  weir,  84 
Triangular  weir,  81 
Tubes,  70 
Turbine  case,  189 
Turbine  pump,  239 


INDEX 

Turbine  setting,  188 
U 

Uniform  flow,  127 
Units,  5 
Unsteady  flow,  148 


Vacuum,  12 

Velocity,  absolute,  152 

diagrams,  211 

critical,  46 

head,  52 

measurement  of,  93 

relative,  152 
Venturi  meter,  74 
Volute  pump,  240 
Vortex,  161,  163 

W 

Water  barometer,  11 
Water  hammer,  107,  148 
Water,  properties  of,  3 
Weirs,  79 
Wetted  perimeter,  128 


267 


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